A. जब आवश्यक पहचान संकेत हट जाएं/When necessary identity cues are removed
Step 1
Concept
Simplicity is useful but identity cues must remain. Exam tip: balance simplification and identity.
Step 2
Why this answer is correct
The correct answer is A. जब आवश्यक पहचान संकेत हट जाएं / When necessary identity cues are removed. Simplicity is useful but identity cues must remain. Exam tip: balance simplification and identity.
Step 3
Exam Tip
सरलता उपयोगी है पर पहचान संकेत बचने चाहिए। परीक्षा में simplification और identity balance रखें।
Reduce factors correctly during simplification. चरण 1: (p=5r) रखने पर \(25r^2=5q^2\) बनता है। चरण 2: दोनों पक्षों को (5) से भाग देने पर \(q^2=5r^2\) मिलता है। चरण 3: सरलीकरण में गुणक सही घटाएँ।
From \(4r^2=2q^2\), dividing both sides by (2) gives \(q^2=2r^2\).
Step 3
Exam Tip
This proves \(q^2\), and then (q), is even. चरण 1: (p=2r) रखने पर \(p^2=4r^2\) होगा। चरण 2: \(4r^2=2q^2\) से दोनों ओर (2) से भाग करने पर \(q^2=2r^2\) मिलता है। चरण 3: इससे \(q^2\) सम और फिर (q) सम सिद्ध होता है।
D. (p=3k) से \(p^2=3k^2\)/From (p=3k), \(p^2=3k^2\)
Step 1
Concept
Squaring (p=3k) gives ((3k)2).
Step 2
Why this answer is correct
Its correct value is \(9k^2\), not \(3k^2\).
Step 3
Exam Tip
Do not forget to square the coefficient. चरण 1: (p=3k) को वर्ग करने पर ((3k)2) मिलता है। चरण 2: इसका सही मान \(9k^2\) है, \(3k^2\) नहीं। चरण 3: गुणांक का वर्ग न भूलें।
Square the coefficient while squaring. चरण 1: (p=2k) है तो (p-2=(2k)2)। चरण 2: इसका सही मान \(4k^2\) है, \(2k^2\) नहीं। चरण 3: वर्ग करते समय गुणांक का भी वर्ग करें।
A. दोनों पक्षों को (3) से भाग देकर \(q^2=3k^2\) पाना/Divide both sides by (3) to get \(q^2=3k^2\)
Step 1
Concept
In \(9k^2=3q^2\), the common factor is (3).
Step 2
Why this answer is correct
Dividing by (3) gives \(3k^2=q^2\), that is \(q^2=3k^2\).
Step 3
Exam Tip
Remove only valid common factors while simplifying. चरण 1: \(9k^2=3q^2\) में साझा गुणनखंड (3) है। चरण 2: (3) से भाग देने पर \(3k^2=q^2\), यानी \(q^2=3k^2\) मिलता है। चरण 3: सरलीकरण में केवल वैध समान गुणनखंड हटाएँ।
In \(25n^2=5y^2\), both sides can be divided by (5).
Step 2
Why this answer is correct
This gives \(5n^2=y^2\), that is \(y^2=5n^2\).
Step 3
Exam Tip
While simplifying, remove only the common factor, not the whole (25). चरण 1: \(25n^2=5y^2\) में दोनों पक्ष (5) से भाग दिए जा सकते हैं। चरण 2: इससे \(5n^2=y^2\), अर्थात \(y^2=5n^2\) मिलता है। चरण 3: सरलीकरण में (25) को पूरा नहीं हटाएँ, केवल समान गुणनखंड हटाएँ।
C. (p=3k) से \(p^2=3k^2\)/From (p=3k), \(p^2=3k^2\)
Step 1
Concept
Squaring (p=3k) gives ((3k)2).
Step 2
Why this answer is correct
The correct value is \(9k^2\), not \(3k^2\).
Step 3
Exam Tip
Square the whole expression. चरण 1: (p=3k) को वर्ग करने पर ((3k)2) मिलेगा। चरण 2: सही मान \(9k^2\) है, \(3k^2\) नहीं। चरण 3: वर्ग करते समय पूरी राशि का वर्ग करें।
A. समांतर श्रेणी है और \(d=\sqrt{3}\)/It is an AP and \(d=\sqrt{3}\)
Step 1
Concept
The terms become \(\sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3}\). In exams, simplify radicals before finding differences.
Step 2
Why this answer is correct
The correct answer is A. समांतर श्रेणी है और \(d=\sqrt{3}\) / It is an AP and \(d=\sqrt{3}\). The terms become \(\sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3}\). In exams, simplify radicals before finding differences.
Step 3
Exam Tip
पद \(\sqrt{3},2\sqrt{3},3\sqrt{3},4\sqrt{3}\) बनते हैं। परीक्षा में मूलों को सरल करके ही अंतर निकालें।
A. समांतर श्रेणी है, \(d=\sqrt{2}\)/It is an AP, \(d=\sqrt{2}\)
Step 1
Concept
The terms become \(\sqrt{2},2\sqrt{2},3\sqrt{2},4\sqrt{2}\), so the difference is \(\sqrt{2}\). In exams, simplify radicals first.
Step 2
Why this answer is correct
The correct answer is A. समांतर श्रेणी है, \(d=\sqrt{2}\) / It is an AP, \(d=\sqrt{2}\). The terms become \(\sqrt{2},2\sqrt{2},3\sqrt{2},4\sqrt{2}\), so the difference is \(\sqrt{2}\). In exams, simplify radicals first.
Step 3
Exam Tip
पद \(\sqrt{2},2\sqrt{2},3\sqrt{2},4\sqrt{2}\) बनते हैं, इसलिए अंतर \(\sqrt{2}\) है। परीक्षा में मूलों को पहले सरल करें।
The first equation becomes (2x+y=19), and substituting (y=x-5) gives (3x-5=19). Simplifying the equation first saves time.
Step 2
Why this answer is correct
The correct answer is D. (x=8,\ y=3). The first equation becomes (2x+y=19), and substituting (y=x-5) gives (3x-5=19). Simplifying the equation first saves time.
Step 3
Exam Tip
पहला समीकरण (2x+y=19) बनता है और (y=x-5) रखने पर (3x-5=19)। समीकरण को पहले सरल करना समय बचाता है।
The first equation becomes (x+y=8); adding it with (x-y=2) gives (2x=10). Simplifying an equation first makes solving easier.
Step 2
Why this answer is correct
The correct answer is C. (x=5,\ y=3). The first equation becomes (x+y=8); adding it with (x-y=2) gives (2x=10). Simplifying an equation first makes solving easier.
Step 3
Exam Tip
पहला समीकरण (x+y=8) बनता है; इसे (x-y=2) से जोड़ने पर (2x=10)। पहले समीकरण को सरल करने से हल आसान होता है।
\( \sqrt{300}=10\sqrt{3} \) and \( \sqrt{147}=7\sqrt{3} \), so the difference is \(3\sqrt{3}\). Simplify the radicals first.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{3}\). \( \sqrt{300}=10\sqrt{3} \) and \( \sqrt{147}=7\sqrt{3} \), so the difference is \(3\sqrt{3}\). Simplify the radicals first.
Step 3
Exam Tip
\( \sqrt{300}=10\sqrt{3} \) और \( \sqrt{147}=7\sqrt{3} \), इसलिए अंतर \(3\sqrt{3}\) है। पहले मूलों को सरल करें।
\( \sqrt{108}=6\sqrt{3} \) and \( \sqrt{48}=4\sqrt{3} \), so the difference is \(2\sqrt{3}\). Subtract only like radicals.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{3}\). \( \sqrt{108}=6\sqrt{3} \) and \( \sqrt{48}=4\sqrt{3} \), so the difference is \(2\sqrt{3}\). Subtract only like radicals.
Step 3
Exam Tip
\( \sqrt{108}=6\sqrt{3} \) और \( \sqrt{48}=4\sqrt{3} \), इसलिए अंतर \(2\sqrt{3}\) है। समान मूलों को ही घटाएँ।
\( \sqrt{192}=8\sqrt{3} \) and \( \sqrt{75}=5\sqrt{3} \), so the difference is \(3\sqrt{3}\). Simplify the radicals first.
Step 2
Why this answer is correct
The correct answer is A. \(3\sqrt{3}\). \( \sqrt{192}=8\sqrt{3} \) and \( \sqrt{75}=5\sqrt{3} \), so the difference is \(3\sqrt{3}\). Simplify the radicals first.
Step 3
Exam Tip
\( \sqrt{192}=8\sqrt{3} \) और \( \sqrt{75}=5\sqrt{3} \), इसलिए अंतर \(3\sqrt{3}\) है। पहले मूलों को सरल करें।
\( \sqrt{75}=5\sqrt{3} \) and \( \sqrt{27}=3\sqrt{3} \), so the difference is \(2\sqrt{3}\). Subtract only like radicals.
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{3}\). \( \sqrt{75}=5\sqrt{3} \) and \( \sqrt{27}=3\sqrt{3} \), so the difference is \(2\sqrt{3}\). Subtract only like radicals.
Step 3
Exam Tip
\( \sqrt{75}=5\sqrt{3} \) और \( \sqrt{27}=3\sqrt{3} \), इसलिए अंतर \(2\sqrt{3}\) है। समान मूलों को ही घटाएँ।
\( \sqrt{12}=2\sqrt{3} \) and \( \sqrt{27}=3\sqrt{3} \), so the sum is \(5\sqrt{3}\). Simplify the radicals first.
Step 2
Why this answer is correct
The correct answer is B. \(5\sqrt{3}\). \( \sqrt{12}=2\sqrt{3} \) and \( \sqrt{27}=3\sqrt{3} \), so the sum is \(5\sqrt{3}\). Simplify the radicals first.
Step 3
Exam Tip
\( \sqrt{12}=2\sqrt{3} \) और \( \sqrt{27}=3\sqrt{3} \), इसलिए योग \(5\sqrt{3}\) है। पहले मूलों को सरल करें।
\( \sqrt{48}=4\sqrt{3} \) and \( \sqrt{27}=3\sqrt{3} \), so the difference is \( \sqrt{3} \). Simplify the radicals first.
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{3} \). \( \sqrt{48}=4\sqrt{3} \) and \( \sqrt{27}=3\sqrt{3} \), so the difference is \( \sqrt{3} \). Simplify the radicals first.
Step 3
Exam Tip
\( \sqrt{48}=4\sqrt{3} \) और \( \sqrt{27}=3\sqrt{3} \), इसलिए अंतर \( \sqrt{3} \) है। पहले मूलों को सरल करें।
\( \sqrt{27}=3\sqrt{3} \) and \( \sqrt{12}=2\sqrt{3} \), so the difference is \( \sqrt{3} \). Subtract like radicals.
Step 2
Why this answer is correct
The correct answer is A. \( \sqrt{3} \). \( \sqrt{27}=3\sqrt{3} \) and \( \sqrt{12}=2\sqrt{3} \), so the difference is \( \sqrt{3} \). Subtract like radicals.
Step 3
Exam Tip
\( \sqrt{27}=3\sqrt{3} \) और \( \sqrt{12}=2\sqrt{3} \) इसलिए अंतर \( \sqrt{3} \) है। समान मूलों को घटाएँ।
Here \(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), and \(3\sqrt{75}=15\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value should be (12).
Step 2
Why this answer is correct
The correct answer is C. (15). Here \(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), and \(3\sqrt{75}=15\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value should be (12).
Step 3
Exam Tip
\(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), और \(3\sqrt{75}=15\sqrt{3}\)। अंश \(12\sqrt{3}\) है, इसलिए मान (12) होना चाहिए।
Here \(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{108}=6\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value is (12).
Step 2
Why this answer is correct
The correct answer is C. (12). Here \(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{108}=6\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value is (12).
Step 3
Exam Tip
\(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), और \(\sqrt{108}=6\sqrt{3}\)। अंश \(12\sqrt{3}\) है, इसलिए मान (12) है।
We have \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). The total is \(4\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(4\sqrt{2}\). We have \(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), and \(\sqrt{72}=6\sqrt{2}\). The total is \(4\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{242}=11\sqrt{2}\), \(\sqrt{128}=8\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), और \(\sqrt{72}=6\sqrt{2}\)। कुल \(4\sqrt{2}\) मिलता है।
Here \(\sqrt{192}=8\sqrt{3}\), \(2\sqrt{48}=8\sqrt{3}\), and \(3\sqrt{12}=6\sqrt{3}\). The numerator is \(6\sqrt{3}\), so the value is (6).
Step 2
Why this answer is correct
The correct answer is C. (12). Here \(\sqrt{192}=8\sqrt{3}\), \(2\sqrt{48}=8\sqrt{3}\), and \(3\sqrt{12}=6\sqrt{3}\). The numerator is \(6\sqrt{3}\), so the value is (6).
Step 3
Exam Tip
\(\sqrt{192}=8\sqrt{3}\), \(2\sqrt{48}=8\sqrt{3}\), और \(3\sqrt{12}=6\sqrt{3}\)। अंश \(6\sqrt{3}\) है, इसलिए मान (6) है।
Here \(\sqrt{108}=6\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{12}=2\sqrt{3}\). The numerator is \(9\sqrt{3}\), so the value is (9).
Step 2
Why this answer is correct
The correct answer is C. (9). Here \(\sqrt{108}=6\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), and \(\sqrt{12}=2\sqrt{3}\). The numerator is \(9\sqrt{3}\), so the value is (9).
Step 3
Exam Tip
\(\sqrt{108}=6\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\), और \(\sqrt{12}=2\sqrt{3}\)। अंश \(9\sqrt{3}\) है, इसलिए मान (9) है।
We have \(\sqrt{162}=9\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\). The total is \(4\sqrt{2}\).
Step 2
Why this answer is correct
The correct answer is C. \(4\sqrt{2}\). We have \(\sqrt{162}=9\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\). The total is \(4\sqrt{2}\).
Step 3
Exam Tip
\(\sqrt{162}=9\sqrt{2}\), \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{50}=5\sqrt{2}\), और \(\sqrt{18}=3\sqrt{2}\)। कुल \(4\sqrt{2}\) मिलता है।
Here \(\sqrt{147}=7\sqrt{3}\), \(2\sqrt{12}=4\sqrt{3}\), and \(3\sqrt{27}=9\sqrt{3}\), so the numerator is \(12\sqrt{3}\). Therefore, the value should be (12).
Step 2
Why this answer is correct
The correct answer is A. (16). Here \(\sqrt{147}=7\sqrt{3}\), \(2\sqrt{12}=4\sqrt{3}\), and \(3\sqrt{27}=9\sqrt{3}\), so the numerator is \(12\sqrt{3}\). Therefore, the value should be (12).
Step 3
Exam Tip
\(\sqrt{147}=7\sqrt{3}\), \(2\sqrt{12}=4\sqrt{3}\), और \(3\sqrt{27}=9\sqrt{3}\), इसलिए अंश \(12\sqrt{3}\) नहीं बल्कि \(7\sqrt{3}-4\sqrt{3}+9\sqrt{3}=12\sqrt{3}\) है। अतः मान (12) होना चाहिए।
Here \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{48}=4\sqrt{3}\), so the numerator is \(9\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (9).
Step 2
Why this answer is correct
The correct answer is A. (9). Here \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{48}=4\sqrt{3}\), so the numerator is \(9\sqrt{3}\). Dividing by \(\sqrt{3}\) gives (9).
Step 3
Exam Tip
\(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{48}=4\sqrt{3}\), इसलिए अंश \(9\sqrt{3}\) है। \(\sqrt{3}\) से भाग देने पर (9) मिलता है।
\(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the numerator is \(\sqrt{5}\), and division gives (1). In exams, first make like radicals.
Step 2
Why this answer is correct
The correct answer is A. (1). \(\sqrt{45}=3\sqrt{5}\) and \(\sqrt{20}=2\sqrt{5}\), so the numerator is \(\sqrt{5}\), and division gives (1). In exams, first make like radicals.
Step 3
Exam Tip
\(\sqrt{45}=3\sqrt{5}\) और \(\sqrt{20}=2\sqrt{5}\), इसलिए ऊपर \(\sqrt{5}\) है और भाग देने पर (1) मिलता है। परीक्षा में पहले समान करणी बनाएं।
Here (\(2^{5}\)^{3}=2^{15}), \(4^{-2}=2^{-4}\), and \(8^{2}=2^{6}\), so the net exponent is (15-4-6=5). In exams, convert all bases to (2).
Step 2
Why this answer is correct
The correct answer is A. \(2^{5}\). Here (\(2^{5}\)^{3}=2^{15}), \(4^{-2}=2^{-4}\), and \(8^{2}=2^{6}\), so the net exponent is (15-4-6=5). In exams, convert all bases to (2).
Step 3
Exam Tip
(\(2^{5}\)^{3}=2^{15}), \(4^{-2}=2^{-4}\), और \(8^{2}=2^{6}\), इसलिए कुल घात (15-4-6=5) है। परीक्षा में सभी आधार (2) में बदलें।
We get \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), and \(\sqrt{8}=2\sqrt{2}\), so the result is \(6\sqrt{2}\). In exams, combine only like surd terms.
Step 2
Why this answer is correct
The correct answer is A. \(6\sqrt{2}\). We get \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), and \(\sqrt{8}=2\sqrt{2}\), so the result is \(6\sqrt{2}\). In exams, combine only like surd terms.
Step 3
Exam Tip
\(\sqrt{50}=5\sqrt{2}\), \(\sqrt{18}=3\sqrt{2}\), और \(\sqrt{8}=2\sqrt{2}\), इसलिए परिणाम \(6\sqrt{2}\) है। परीक्षा में समान करणी पदों को ही जोड़ें।
Inside, \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), and raising to (-2) gives \(\frac{y^{8}}{4x^{6}}\). In exams, simplify inside the bracket first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{y^{8}}{4x^{6}}\). Inside, \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), and raising to (-2) gives \(\frac{y^{8}}{4x^{6}}\). In exams, simplify inside the bracket first.
Step 3
Exam Tip
अंदर \(\frac{4x^{2}y^{-3}}{2x^{-1}y}=2x^{3}y^{-4}\), इसलिए घात (-2) देने पर \(\frac{y^{8}}{4x^{6}}\) मिलता है। परीक्षा में पहले कोष्ठक के अंदर सरल करें।
The numerator exponent is ((m+2)+(3-m)=5), and \(\frac{a^{5}}{a^{4}}=a\). In exams, add and subtract exponents only for the same base.
Step 2
Why this answer is correct
The correct answer is A. (a). The numerator exponent is ((m+2)+(3-m)=5), and \(\frac{a^{5}}{a^{4}}=a\). In exams, add and subtract exponents only for the same base.
Step 3
Exam Tip
ऊपर की घातें ((m+2)+(3-m)=5) हैं और \(\frac{a^{5}}{a^{4}}=a\)। परीक्षा में समान आधार की घातों को जोड़ना और घटाना याद रखें।
\(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{50}=5\sqrt{2}\), so the answer is \(8\sqrt{2}\). In exams, first write all surds in simplest form.
Step 2
Why this answer is correct
The correct answer is A. \(,8\sqrt{2},\). \(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), and \(\sqrt{50}=5\sqrt{2}\), so the answer is \(8\sqrt{2}\). In exams, first write all surds in simplest form.
Step 3
Exam Tip
\(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\) और \(\sqrt{50}=5\sqrt{2}\), इसलिए उत्तर \(8\sqrt{2}\) है। परीक्षा में पहले सभी surds को simplest form में लिखें।
The numerator difference is (6x-2y+2y-3=2y\(3x^2+y^2\)), so division gives \(3x^2+y^2\). In exams, take out the common factor.
Step 2
Why this answer is correct
The correct answer is A. \(,3x^2+y^2,\). The numerator difference is (6x-2y+2y-3=2y\(3x^2+y^2\)), so division gives \(3x^2+y^2\). In exams, take out the common factor.
Step 3
Exam Tip
ऊपर का अंतर (6x-2y+2y-3=2y\(3x^2+y^2\)) है, इसलिए भाग देने पर \(3x^2+y^2\) मिलता है। परीक्षा में common factor निकालें।
Because \(\sqrt{a^4}=a^2\) and \(\sqrt{b^2}=b\), the simplified form is \(a^2b\). In exams, note the positive condition.
Step 2
Why this answer is correct
The correct answer is A. \(,a^2b,\). Because \(\sqrt{a^4}=a^2\) and \(\sqrt{b^2}=b\), the simplified form is \(a^2b\). In exams, note the positive condition.
Step 3
Exam Tip
क्योंकि \(\sqrt{a^4}=a^2\) और \(\sqrt{b^2}=b\), इसलिए सरल रूप \(a^2b\) है। परीक्षा में positive condition को ध्यान में रखें।
(x-4-81=\(x^2-9\)\(x^2+9\)), so the simplified form is \(x^2-9\). In exams, treat \(x^4\) as (\(x^2\)2) while factoring.
Step 2
Why this answer is correct
The correct answer is A. \(,x^2-9,\). (x-4-81=\(x^2-9\)\(x^2+9\)), so the simplified form is \(x^2-9\). In exams, treat \(x^4\) as (\(x^2\)2) while factoring.
Step 3
Exam Tip
(x-4-81=\(x^2-9\)\(x^2+9\)), इसलिए सरल रूप \(x^2-9\) है। परीक्षा में \(x^4\) को (\(x^2\)2) मानकर factor करें।
This is of the form ((A+B)2-(A-B)2=4AB), where (A=3x) and (B=2), so the answer is (24x). In exams, identities save time.
Step 2
Why this answer is correct
The correct answer is A. (,24x,). This is of the form ((A+B)2-(A-B)2=4AB), where (A=3x) and (B=2), so the answer is (24x). In exams, identities save time.
Step 3
Exam Tip
यह ((A+B)2-(A-B)2=4AB) का रूप है, जहां (A=3x) और (B=2), इसलिए उत्तर (24x) है। परीक्षा में identity से समय बचता है।
\(\sqrt{75}=5\sqrt{3}\), \(\sqrt{12}=2\sqrt{3}\), and \(\sqrt{48}=4\sqrt{3}\), so the answer is \(7\sqrt{3}\). In exams, combine only terms with the same radical part.
Step 2
Why this answer is correct
The correct answer is A. \(,7\sqrt{3},\). \(\sqrt{75}=5\sqrt{3}\), \(\sqrt{12}=2\sqrt{3}\), and \(\sqrt{48}=4\sqrt{3}\), so the answer is \(7\sqrt{3}\). In exams, combine only terms with the same radical part.
Step 3
Exam Tip
\(\sqrt{75}=5\sqrt{3}\), \(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{48}=4\sqrt{3}\), इसलिए उत्तर \(7\sqrt{3}\) है। परीक्षा में समान मूल वाले पद ही जोड़ें।
The numerator is (\(a^{-2}b^3\)2=a^{-4}b-6) and the denominator is (\(ab^{-1}\)^{-1}=a^{-1}b), so the answer is \(\dfrac{b^5}{a^3}\). In exams, apply the outside power first.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{b^5}{a^3},\). The numerator is (\(a^{-2}b^3\)2=a^{-4}b-6) and the denominator is (\(ab^{-1}\)^{-1}=a^{-1}b), so the answer is \(\dfrac{b^5}{a^3}\). In exams, apply the outside power first.
Step 3
Exam Tip
ऊपर (\(a^{-2}b^3\)2=a^{-4}b-6) और नीचे (\(ab^{-1}\)^{-1}=a^{-1}b), इसलिए उत्तर \(\dfrac{b^5}{a^3}\) है। परीक्षा में बाहर की घात पहले लगाएं।
The expression inside is \(a^{-3}b^4\), and the power (-1) gives its reciprocal \(\dfrac{a^3}{b^4}\). In exams, apply the outer negative power at the end.
Step 2
Why this answer is correct
The correct answer is A. \(,\dfrac{a^3}{b^4},\). The expression inside is \(a^{-3}b^4\), and the power (-1) gives its reciprocal \(\dfrac{a^3}{b^4}\). In exams, apply the outer negative power at the end.
Step 3
Exam Tip
अंदर का भाग \(a^{-3}b^4\) है, और (-1) घात से उसका व्युत्क्रम \(\dfrac{a^3}{b^4}\) हो जाता है। परीक्षा में outer negative power अंत में लगाएं।
(x-4-16=\(x^2-4\)\(x^2+4\)), so the simplified form is \(x^2+4\). In exams, treat \(x^4\) as (\(x^2\)2) for factorisation.
Step 2
Why this answer is correct
The correct answer is A. \(,x^2+4,\). (x-4-16=\(x^2-4\)\(x^2+4\)), so the simplified form is \(x^2+4\). In exams, treat \(x^4\) as (\(x^2\)2) for factorisation.
Step 3
Exam Tip
(x-4-16=\(x^2-4\)\(x^2+4\)), इसलिए सरल रूप \(x^2+4\) है। परीक्षा में \(x^4\) को (\(x^2\)2) समझकर factor करें।
Because \(\sqrt{a^2}=a\) and \(\sqrt{b^4}=b^2\), the answer is \(ab^2\). In exams, note the condition that variables are positive.
Step 2
Why this answer is correct
The correct answer is A. \(,ab^2,\). Because \(\sqrt{a^2}=a\) and \(\sqrt{b^4}=b^2\), the answer is \(ab^2\). In exams, note the condition that variables are positive.
Step 3
Exam Tip
क्योंकि \(\sqrt{a^2}=a\) और \(\sqrt{b^4}=b^2\), इसलिए उत्तर \(ab^2\) है। परीक्षा में variables के positive होने की शर्त ध्यान रखें।
On expansion, ((p+q)2=p-2+2pq+q-2) and ((p-q)2=p-2-2pq+q-2), so the difference is (4pq). In exams, apply standard identities directly.
Step 2
Why this answer is correct
The correct answer is A. (,4pq,). On expansion, ((p+q)2=p-2+2pq+q-2) and ((p-q)2=p-2-2pq+q-2), so the difference is (4pq). In exams, apply standard identities directly.
Step 3
Exam Tip
विस्तार करने पर ((p+q)2=p-2+2pq+q-2) और ((p-q)2=p-2-2pq+q-2), इसलिए अंतर (4pq) है। परीक्षा में standard identities सीधे लगाएं।
Inside, \(a^{\frac{1}{2}}a^{\frac{3}{2}}=a^2\), so (\dfrac{\(a^2\)2}{a-3}=a). In exams, solve fractional exponents using the usual exponent rules.
Step 2
Why this answer is correct
The correct answer is A. (,a,). Inside, \(a^{\frac{1}{2}}a^{\frac{3}{2}}=a^2\), so (\dfrac{\(a^2\)2}{a-3}=a). In exams, solve fractional exponents using the usual exponent rules.
Step 3
Exam Tip
अंदर \(a^{\frac{1}{2}}a^{\frac{3}{2}}=a^2\), इसलिए (\dfrac{\(a^2\)2}{a-3}=a)। परीक्षा में fractional exponents को भी सामान्य घात नियम से हल करें।
Because \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{8}=2\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\), the answer is \(4\sqrt{2}\). In exams, combine only like surd terms.
Step 2
Why this answer is correct
The correct answer is A. \(,4\sqrt{2},\). Because \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{8}=2\sqrt{2}\), and \(\sqrt{18}=3\sqrt{2}\), the answer is \(4\sqrt{2}\). In exams, combine only like surd terms.
Step 3
Exam Tip
क्योंकि \(\sqrt{50}=5\sqrt{2}\), \(\sqrt{8}=2\sqrt{2}\) और \(\sqrt{18}=3\sqrt{2}\), इसलिए उत्तर \(4\sqrt{2}\) है। परीक्षा में समान surd terms को ही जोड़ें या घटाएं।
The numerator gives \(a^m \times a^{2m}=a^{3m}\), and then \(\dfrac{a^{3m}}{a^{3m-2}}=a^2\). In exams, subtract exponents during division.
Step 2
Why this answer is correct
The correct answer is A. \(,a^2,\). The numerator gives \(a^m \times a^{2m}=a^{3m}\), and then \(\dfrac{a^{3m}}{a^{3m-2}}=a^2\). In exams, subtract exponents during division.
Step 3
Exam Tip
ऊपर \(a^m \times a^{2m}=a^{3m}\) और फिर \(\dfrac{a^{3m}}{a^{3m-2}}=a^2\) होगा। परीक्षा में भाग करते समय घातांक घटाएं।
First write (43=\(2^2\)3=26) and (82=\(2^3\)2=26). Thus \(\frac{2^3\cdot2^6}{2^6}=2^{3+6-6}=2^3\), so the correct option is \(2^3\).
Step 2
Why this answer is correct
The correct answer is C. \(2^5\). First write (43=\(2^2\)3=26) and (82=\(2^3\)2=26). Thus \(\frac{2^3\cdot2^6}{2^6}=2^{3+6-6}=2^3\), so the correct option is \(2^3\).
Step 3
Exam Tip
पहले (43=\(2^2\)3=26) और (82=\(2^3\)2=26) लिखें। इसलिए \(\frac{2^3\cdot2^6}{2^6}=2^3\) नहीं बल्कि \(2^{3+6-6}=2^3\); सही विकल्प \(2^3\) है।
Dividing the whole equation by (49) gives (x-2-(r+s)x+rs=0). In exams, removing the common factor first shortens the solution.
Step 2
Why this answer is correct
The correct answer is A. (x=r,s). Dividing the whole equation by (49) gives (x-2-(r+s)x+rs=0). In exams, removing the common factor first shortens the solution.
Step 3
Exam Tip
पूरे समीकरण को (49) से भाग देने पर (x-2-(r+s)x+rs=0) मिलता है। परीक्षा में पहले सामान्य गुणक हटाना हल को छोटा करता है।
Dividing the whole equation by (36) gives (x-2-(m+n)x+mn=0). In exams, removing the common factor first shortens the solution.
Step 2
Why this answer is correct
The correct answer is A. (x=m,n). Dividing the whole equation by (36) gives (x-2-(m+n)x+mn=0). In exams, removing the common factor first shortens the solution.
Step 3
Exam Tip
पूरे समीकरण को (36) से भाग देने पर (x-2-(m+n)x+mn=0) मिलता है। परीक्षा में पहले सामान्य गुणक हटाना हल को छोटा करता है।
Dividing the whole equation by (25) gives (x-2-(a+b)x+ab=0). In exams, removing the common factor first shortens the solution.
Step 2
Why this answer is correct
The correct answer is A. (x=a,b). Dividing the whole equation by (25) gives (x-2-(a+b)x+ab=0). In exams, removing the common factor first shortens the solution.
Step 3
Exam Tip
पूरे समीकरण को (25) से भाग देने पर (x-2-(a+b)x+ab=0) मिलता है। परीक्षा में पहले सामान्य गुणक हटाना हल को छोटा करता है।
Dividing the whole equation by (16) gives (x-2-(a+b)x+ab=0). In exams, removing the common factor first makes solving easier.
Step 2
Why this answer is correct
The correct answer is A. (x=a,b). Dividing the whole equation by (16) gives (x-2-(a+b)x+ab=0). In exams, removing the common factor first makes solving easier.
Step 3
Exam Tip
पूरे समीकरण को (16) से भाग देने पर (x-2-(a+b)x+ab=0) मिलता है। परीक्षा में पहले सामान्य गुणक हटाना हल को आसान करता है।
Dividing the whole equation by (9) gives (x-2-(r+s)x+rs=0). In exams, removing the common factor first makes solving easier.
Step 2
Why this answer is correct
The correct answer is A. (x=r,s). Dividing the whole equation by (9) gives (x-2-(r+s)x+rs=0). In exams, removing the common factor first makes solving easier.
Step 3
Exam Tip
पूरे समीकरण को (9) से भाग देने पर (x-2-(r+s)x+rs=0) मिलता है। परीक्षा में पहले सामान्य गुणक हटाने से हल आसान होता है।
\(\frac{9}{\sqrt{2}}\) simplifies to \(\frac{9\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 2
Why this answer is correct
The correct answer is A. \( \frac{9\sqrt{2}}{2}\). \(\frac{9}{\sqrt{2}}\) simplifies to \(\frac{9\sqrt{2}}{2}\). In exams, do not forget to rationalize the denominator.
Step 3
Exam Tip
\(\frac{9}{\sqrt{2}}\) को सरल करने पर \(\frac{9\sqrt{2}}{2}\) मिलता है। परीक्षा में हर को परिमेय बनाना न भूलें।
Dividing every term by (8) gives \(x^2-4x+3=0\). Dividing by a common nonzero factor does not change the roots.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-4x+3=0\). Dividing every term by (8) gives \(x^2-4x+3=0\). Dividing by a common nonzero factor does not change the roots.
Step 3
Exam Tip
हर पद को (8) से भाग देने पर \(x^2-4x+3=0\) मिलता है। समान अशून्य गुणनखंड से भाग देने पर मूल नहीं बदलते।
