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।
(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।
(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।
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।
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।
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।
