Question

please solve these questions fastly

Answer 1

Identities used :

${(x + y)}^{2} = {x}^{2} + {y}^{2} + 2xy \\ \\ {(x - y)}^{2} = {x}^{2} + {y}^{2} - 2xy$

$3. \: (i) \: \sqrt{7} ( \sqrt{35} + \sqrt{7} ) \\ \\ = \sqrt{35 \times 7} + {( \sqrt{7} )}^{2} \\ \\ = \sqrt{ {7}^{2} \times 5 } + 7 \\ \\ = 7 \sqrt{5} + 7$
$(ii) \: {( \sqrt{x} - 5 )}^{2} \\ \\ = {( \sqrt{x} )}^{2} + {(5)}^{2} + 2( \sqrt{x} )(5) \\ \\ = x + 25 + 10 \sqrt{x}$

$4. \: 4 \sqrt[3]{40} + 3 \sqrt[3]{625} - 4 \sqrt[3]{320} \\ \\ = 4 \sqrt[3]{ {2}^{3} \times 5} + 3 \sqrt[3]{ {5}^{3} } - 4 \sqrt[3]{ {4}^{3} \times 5 } \\ \\ = 4 \times 2 \sqrt[3]{5} + 3 \times 5 - 4 \times 4 \sqrt[3]{5} \\ \\ = 8 \sqrt[3]{5} + 15 - 16 \sqrt[3]{5} \\ \\ = 15 + 8 \sqrt[3]{5} - 16 \sqrt[3]{5} \\ \\ = 15 - 8 \sqrt[3]{5}$

$5. \: \frac{70}{ \sqrt{10} + \sqrt{20} + \sqrt{40} - \sqrt{80} } \\ \\ \bf \: Solving \: the \: given \: equation \\ \bf \: we \: get : \\ \\ = \frac{70}{ \sqrt{10} + \sqrt{ {2}^{2} \times 5} + \sqrt{ {2}^{2} \times 10 } - \sqrt{ {4}^{2} \times 5 } } \\ \\ = \frac{70}{ \sqrt{10} + 2 \sqrt{5} + 2 \sqrt{10} - 4\sqrt{5} } \\ \\ = \frac{70}{ 3\sqrt{10} - 2 \sqrt{5} } \\ \\ \bf \: On \: rationalising \: the \: denominator \\ \bf \: we \: get \\ \\ = \frac{70}{3 \sqrt{10} - 2 \sqrt{5} } \times \frac{3 \sqrt{10} + 2 \sqrt{5} }{3 \sqrt{10} + 2 \sqrt{5} } \\ \\ = \frac{210 \sqrt{10} - 140 \sqrt{5} }{ {(3 \sqrt{10} )}^{2} - {(2 \sqrt{5}) }^{2} } \\ \\ = \frac{210 \sqrt{10} - 140 \sqrt{5} }{90 - 20} \\ \\ = \frac{210 \sqrt{10} - 140 \sqrt{5} }{70} \\ \\ = \frac{70(3 \sqrt{10} - 2 \sqrt{5}) }{70} \\ \\ = 3 \sqrt{10} - 2 \sqrt{5}$

$6. \bf \: Given \: \\ x = 3 - 2 \sqrt{2} \\ \\ (i) \frac{1}{x} = \frac{1}{3 - 2 \sqrt{2} } \\ \\ \bf \: On \: rationalising \: the \: denominator \: we \: get \\ \\ = \frac{1}{3 - 2 \sqrt{2} } \times \frac{3 + 2 \sqrt{2} }{3 + 2 \sqrt{2} } \\ \\ = \frac{3 + 2 \sqrt{2} }{ {(3)}^{2} - {(2 \sqrt{2} )}^{2} } \\ \\ = \frac{3 + 2 \sqrt{2} }{9 - 8} \\ \\ = 3 + 2 \sqrt{2} \\ \\ (ii) \: x + \frac{1}{x} = (3 - 2\sqrt{2} ) + (3 + 2 \sqrt{2} ) \\ \\ = 3 - 2 \sqrt{2} + 3 + 2 \sqrt{2} \\ \\ = 3 + 3 \\ \\ = 6 \\ \\ (iii) \: x - \frac{1}{x} = (3 - 2 \sqrt{2} ) - (3 + 2 \sqrt{2} \\ \\ = 3 - 2 \sqrt{2} - 3 - 2 \sqrt{2} \\ \\ = - 2 \sqrt{2} - 2 \sqrt{2} \\ \\ = - 4 \sqrt{2}$

$7. \: \: \bf \: Given \\ a = 7 - 4 \sqrt{3} \\ \\ \frac{1}{a} = \frac{1}{7 - 4 \sqrt{3} } \\ \\ \frac{1}{a} = \frac{1}{7 - 4 \sqrt{3} } \times \frac{7 + 4 \sqrt{3} }{7 + 4 \sqrt{3} } \\ \\ \frac{1}{a} = \frac{7 + 4 \sqrt{3} }{ {(7)}^{2} - {4 \sqrt{3} )}^{2} } \\ \\ \frac{1}{a} = \frac{7 + 4 \sqrt{3} }{49 - 48} \\ \\ \frac{1}{a} = 7 + 4 \sqrt{3} \\ \\ \sqrt{a} + \frac{1}{ \sqrt{a} } \\ \\ = \sqrt{7 - 4 \sqrt{3} } + \sqrt{7 + 4 \sqrt{3} } \\ \\ \bf \: we \: can \: write \: it \: as \\ \\ \sqrt{ {(2)}^{2} + {( \sqrt{3}) }^{2} - 2 (2)( \sqrt{3}) } \\ + \sqrt{ {(2)}^{2} + {( \sqrt{3}) }^{2} - 2(2)( \sqrt{3}) } \\ \\ = \sqrt{ {(2 - \sqrt{3} )}^{2} } + \sqrt{ {(2 + \sqrt{3} )}^{2} } \\ \\ = (2 - \sqrt{3} ) + (2 + \sqrt{3} ) \\ \\ = 2 - \sqrt{3} + 2 + \sqrt{3} \\ \\ = 2 + 2 \\ \\ = 4$