Here (u^{2}-v^{2}=(u-v)(u+v)=4\sqrt{3}\cdot2\sqrt{7}=8\sqrt{21}) and (uv=4). Therefore, the value is \(2\sqrt{21}\).
Step 2
Why this answer is correct
The correct answer is A. \(2\sqrt{21}\). Here (u^{2}-v^{2}=(u-v)(u+v)=4\sqrt{3}\cdot2\sqrt{7}=8\sqrt{21}) and (uv=4). Therefore, the value is \(2\sqrt{21}\).
Step 3
Exam Tip
यहाँ (u^{2}-v^{2}=(u-v)(u+v)=4\sqrt{3}\cdot2\sqrt{7}=8\sqrt{21}) और (uv=4) है। इसलिए मान \(2\sqrt{21}\) है।
(\(\sqrt{3}-\sqrt{2}\)^{2}=3+2-2\sqrt{6}=5-2\sqrt{6}). In exams, identify (a,b) from (a+b) and \(2\sqrt{ab}\).
Step 2
Why this answer is correct
The correct answer is A. \((\sqrt{3}-\sqrt{2})^{2}\). (\(\sqrt{3}-\sqrt{2}\)^{2}=3+2-2\sqrt{6}=5-2\sqrt{6}). In exams, identify (a,b) from (a+b) and \(2\sqrt{ab}\).
Step 3
Exam Tip
(\(\sqrt{3}-\sqrt{2}\)^{2}=3+2-2\sqrt{6}=5-2\sqrt{6})। परीक्षा में (a+b) और \(2\sqrt{ab}\) से (a,b) पहचानें।
Since (\(2+\sqrt{3}\)^{2}=4+3+4\sqrt{3}=7+4\sqrt{3}), \(\sqrt{A}=2+\sqrt{3}\). In exams, recognize the form ((a+b)^{2}).
Step 2
Why this answer is correct
The correct answer is A. \(2+\sqrt{3}\). Since (\(2+\sqrt{3}\)^{2}=4+3+4\sqrt{3}=7+4\sqrt{3}), \(\sqrt{A}=2+\sqrt{3}\). In exams, recognize the form ((a+b)^{2}).
Step 3
Exam Tip
(\(2+\sqrt{3}\)^{2}=4+3+4\sqrt{3}=7+4\sqrt{3}), इसलिए \(\sqrt{A}=2+\sqrt{3}\)। परीक्षा में रूप ((a+b)^{2}) पहचानें।
(\(\sqrt{11}-\sqrt{2}\)^{2}=11+2-2\sqrt{22}=13-2\sqrt{22}). In exams, include both \(+b^{2}\) and (-2ab) in ((a-b)^{2}).
Step 2
Why this answer is correct
The correct answer is A. \(13-2\sqrt{22}\). (\(\sqrt{11}-\sqrt{2}\)^{2}=11+2-2\sqrt{22}=13-2\sqrt{22}). In exams, include both \(+b^{2}\) and (-2ab) in ((a-b)^{2}).
Step 3
Exam Tip
(\(\sqrt{11}-\sqrt{2}\)^{2}=11+2-2\sqrt{22}=13-2\sqrt{22})। परीक्षा में ((a-b)^{2}) में \(+b^{2}\) और (-2ab) दोनों लिखें।
(\(\sqrt{6}+\sqrt{2}\)^{2}=6+2+2\sqrt{12}=8+4\sqrt{3}). In exams, do not miss the middle term of ((a+b)^{2}).
Step 2
Why this answer is correct
The correct answer is A. \(8+4\sqrt{3}\). (\(\sqrt{6}+\sqrt{2}\)^{2}=6+2+2\sqrt{12}=8+4\sqrt{3}). In exams, do not miss the middle term of ((a+b)^{2}).
Step 3
Exam Tip
(\(\sqrt{6}+\sqrt{2}\)^{2}=6+2+2\sqrt{12}=8+4\sqrt{3})। परीक्षा में ((a+b)^{2}) का मध्य पद न भूलें।
Rationalizing gives \(\frac{1}{\sqrt{5}+2}\cdot\frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2\). In exams, use the conjugate of the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}-2\). Rationalizing gives \(\frac{1}{\sqrt{5}+2}\cdot\frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2\). In exams, use the conjugate of the denominator.
Step 3
Exam Tip
\(\frac{1}{\sqrt{5}+2}\cdot\frac{\sqrt{5}-2}{\sqrt{5}-2}=\frac{\sqrt{5}-2}{5-4}=\sqrt{5}-2\)। परीक्षा में हर के संयुग्म का प्रयोग करें।
Multiplying by \(2+\sqrt{3}\) makes the denominator (4-3=1). In exams, multiply both numerator and denominator by the conjugate.
Step 2
Why this answer is correct
The correct answer is A. \(,6+3\sqrt{3},\). Multiplying by \(2+\sqrt{3}\) makes the denominator (4-3=1). In exams, multiply both numerator and denominator by the conjugate.
Step 3
Exam Tip
हर को \(2+\sqrt{3}\) से गुणा करने पर हर (4-3=1) हो जाता है। परीक्षा में conjugate से numerator और denominator दोनों को गुणा करें।
\(\sqrt{98}=7\sqrt{2}\), \(\sqrt{72}=6\sqrt{2}\), 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।
Since \(\sqrt{8}=2\sqrt{2}\), (\(\sqrt{2}+\sqrt{8}\)2=\(3\sqrt{2}\)2=18). In exams, simplify the surd before squaring.
Step 2
Why this answer is correct
The correct answer is A. (,18,). Since \(\sqrt{8}=2\sqrt{2}\), (\(\sqrt{2}+\sqrt{8}\)2=\(3\sqrt{2}\)2=18). In exams, simplify the surd before squaring.
Step 3
Exam Tip
क्योंकि \(\sqrt{8}=2\sqrt{2}\), इसलिए (\(\sqrt{2}+\sqrt{8}\)2=\(3\sqrt{2}\)2=18)। परीक्षा में वर्ग करने से पहले surd सरल करें।
\(\dfrac{\sqrt{48}}{\sqrt{3}}=\sqrt{16}=4\) and \(\dfrac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\), so the sum is (9). In exams, simplify the division inside the root.
Step 2
Why this answer is correct
The correct answer is A. (,9,). \(\dfrac{\sqrt{48}}{\sqrt{3}}=\sqrt{16}=4\) and \(\dfrac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\), so the sum is (9). In exams, simplify the division inside the root.
Step 3
Exam Tip
\(\dfrac{\sqrt{48}}{\sqrt{3}}=\sqrt{16}=4\) और \(\dfrac{\sqrt{75}}{\sqrt{3}}=\sqrt{25}=5\), इसलिए योग (9) है। परीक्षा में root के अंदर भाग को सरल करें।
\(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the inside value is \(-\sqrt{3}\) and the product is (-6). In exams, simplify the surds first.
Step 2
Why this answer is correct
The correct answer is A. (,-6,). \(\sqrt{12}=2\sqrt{3}\) and \(\sqrt{27}=3\sqrt{3}\), so the inside value is \(-\sqrt{3}\) and the product is (-6). In exams, simplify the surds first.
Step 3
Exam Tip
\(\sqrt{12}=2\sqrt{3}\) और \(\sqrt{27}=3\sqrt{3}\), इसलिए अंदर का मान \(-\sqrt{3}\) है और गुणनफल (-6) है। परीक्षा में पहले surd को सरल करें।
Multiplying by \(\sqrt{7}-\sqrt{5}\) makes the denominator (7-5=2) and gives \(\sqrt{7}-\sqrt{5}\). In exams, use the conjugate.
Step 2
Why this answer is correct
The correct answer is A. \(,\sqrt{7}-\sqrt{5},\). Multiplying by \(\sqrt{7}-\sqrt{5}\) makes the denominator (7-5=2) and gives \(\sqrt{7}-\sqrt{5}\). In exams, use the conjugate.
Step 3
Exam Tip
हर को \(\sqrt{7}-\sqrt{5}\) से गुणा करने पर हर (7-5=2) होता है और उत्तर \(\sqrt{7}-\sqrt{5}\) मिलता है। परीक्षा में conjugate का प्रयोग करें।
\(\sqrt{75}=5\sqrt{3}\), \(\sqrt{12}=2\sqrt{3}\), 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।
When the two squares are added, the surd terms cancel and (7+7=14). In exams, irrational terms often cancel in conjugate expressions.
Step 2
Why this answer is correct
The correct answer is A. (,14,). When the two squares are added, the surd terms cancel and (7+7=14). In exams, irrational terms often cancel in conjugate expressions.
Step 3
Exam Tip
दोनों वर्ग जोड़ने पर surd terms कट जाते हैं और (7+7=14) मिलता है। परीक्षा में conjugate expressions में irrational terms अक्सर cancel होते हैं।
Multiplying by \(\sqrt{3}+\sqrt{2}\) makes the denominator (3-2=1). In exams, remember to multiply by the conjugate.
Step 2
Why this answer is correct
The correct answer is A. \(,\sqrt{3}+\sqrt{2},\). Multiplying by \(\sqrt{3}+\sqrt{2}\) makes the denominator (3-2=1). In exams, remember to multiply by the conjugate.
Step 3
Exam Tip
हर को \(\sqrt{3}+\sqrt{2}\) से गुणा करने पर हर (3-2=1) हो जाता है। परीक्षा में conjugate से गुणा करना न भूलें।
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।
\(\frac{1}{2+\sqrt{7}}=\frac{\sqrt{7}-2}{3}\) so the total is \(\frac{4+4\sqrt{7}}{3}\). In exams rationalize the reciprocal first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4+4\sqrt{7}}{3}\). \(\frac{1}{2+\sqrt{7}}=\frac{\sqrt{7}-2}{3}\) so the total is \(\frac{4+4\sqrt{7}}{3}\). In exams rationalize the reciprocal first.
Step 3
Exam Tip
\(\frac{1}{2+\sqrt{7}}=\frac{\sqrt{7}-2}{3}\) है इसलिए कुल \(\frac{4+4\sqrt{7}}{3}\) मिलता है। परीक्षा में व्युत्क्रम को पहले परिमेयकृत करें।
The conjugate of the denominator is \(\sqrt{13}+2\) and the denominator becomes (13-4=9). Hence the value is (\frac{3\(\sqrt{13}+2\)}{9}=\frac{\sqrt{13}+2}{3}).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{13}+2}{3}\). The conjugate of the denominator is \(\sqrt{13}+2\) and the denominator becomes (13-4=9). Hence the value is (\frac{3\(\sqrt{13}+2\)}{9}=\frac{\sqrt{13}+2}{3}).
Step 3
Exam Tip
हर का संयुग्मी \(\sqrt{13}+2\) है और हर (13-4=9) बनता है। इसलिए मान (\frac{3\(\sqrt{13}+2\)}{9}=\frac{\sqrt{13}+2}{3}) है।
(\(\sqrt{28}\)\(\sqrt{7}\)=\sqrt{196}=14) which is rational. In exams keep multiplication and addition rules separate.
Step 2
Why this answer is correct
The correct answer is A. (\(\sqrt{28}\)\(\sqrt{7}\)). (\(\sqrt{28}\)\(\sqrt{7}\)=\sqrt{196}=14) which is rational. In exams keep multiplication and addition rules separate.
Step 3
Exam Tip
(\(\sqrt{28}\)\(\sqrt{7}\)=\sqrt{196}=14) है जो परिमेय है। परीक्षा में गुणन और जोड़ के नियम अलग रखें।
Multiplying by the conjugate \(\sqrt{11}+3\) makes the denominator (11-9=2), and (2) cancels. In exams choose the conjugate of the denominator correctly.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{11}+3\). Multiplying by the conjugate \(\sqrt{11}+3\) makes the denominator (11-9=2), and (2) cancels. In exams choose the conjugate of the denominator correctly.
Step 3
Exam Tip
हर के संयुग्मी \(\sqrt{11}+3\) से गुणा करने पर हर (11-9=2) बनता है और (2) कट जाता है। परीक्षा में हर का संयुग्मी सही चुनें।
The conjugate of the denominator is \(\sqrt{7}-\sqrt{6}\), and the denominator becomes (7-6=1). In exams the answer simplifies when the difference is (1).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{7}-\sqrt{6}\). The conjugate of the denominator is \(\sqrt{7}-\sqrt{6}\), and the denominator becomes (7-6=1). In exams the answer simplifies when the difference is (1).
Step 3
Exam Tip
हर का संयुग्मी \(\sqrt{7}-\sqrt{6}\) है और हर (7-6=1) बनता है। परीक्षा में अंतर (1) होने पर उत्तर सरल हो जाता है।
(x-2-y-2=(x-y)(x+y)=\(2\sqrt{2}\)\(2\sqrt{5}\)=4\sqrt{10}). In exams use identities to avoid long calculation.
Step 2
Why this answer is correct
The correct answer is A. \(4\sqrt{10}\). (x-2-y-2=(x-y)(x+y)=\(2\sqrt{2}\)\(2\sqrt{5}\)=4\sqrt{10}). In exams use identities to avoid long calculation.
Step 3
Exam Tip
(x-2-y-2=(x-y)(x+y)=\(2\sqrt{2}\)\(2\sqrt{5}\)=4\sqrt{10}) है। परीक्षा में पहचान का प्रयोग करके लंबी गणना बचाएं।
Rationalizing \(\frac{1}{2-\sqrt{3}}\) with \(2+\sqrt{3}\) gives \(2+\sqrt{3}\). In exams multiply by the conjugate of the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(2+\sqrt{3}\). Rationalizing \(\frac{1}{2-\sqrt{3}}\) with \(2+\sqrt{3}\) gives \(2+\sqrt{3}\). In exams multiply by the conjugate of the denominator.
Step 3
Exam Tip
\(\frac{1}{2-\sqrt{3}}\) को \(2+\sqrt{3}\) से परिमेयकृत करने पर \(2+\sqrt{3}\) मिलता है। परीक्षा में हर का संयुग्मी लगाएं।
Multiplying the denominator by \(\sqrt{5}+2\) makes it (5-4=1). In exams choose the conjugate of the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{5}+2\). Multiplying the denominator by \(\sqrt{5}+2\) makes it (5-4=1). In exams choose the conjugate of the denominator.
Step 3
Exam Tip
हर को \(\sqrt{5}+2\) से गुणा करने पर हर (5-4=1) बनता है। परीक्षा में हर का संयुग्मी चुनें।
