Question

2. Solve:
a) (V2 + 5) (V2 - 6)
-1
-1
b) [(27)3 x (81) 2 ] = [(36)6 (36)
Given:Fig​

Answer 1

Answer:

a)

$(\sqrt{2} + 5)(\sqrt{2} - 6)\\\\= \sqrt{2}(\sqrt{2} - 6) + 5(\sqrt{2} - 6)\\\\= 2 - 6\sqrt{2} + 5 \sqrt{2} - 30\\\\= -28 - \sqrt{2}$

b)

$[(27)^{\frac{1}{3}} * (81)^{\frac{-1}{2}}]$ ÷ $[(36)^{\frac{1}{6}} * (36)^{\frac{-1}{3}}]$

$= [(3^{3})^{\frac{1}{3}} * (9^{2})^{\frac{-1}{2}}]$ ÷ $[(6^{2} )^{\frac{1}{6}} * ( 6^{2})^{\frac{-1}{3}}]$

$= [(3) * (9)^{-1}]$ ÷ $[(6)^{\frac{1}{3}} * ( 6)^{\frac{-2}{3}}]$

$= [(3) * (3^{2})^{-1}]$ ÷ $[(6)^{\frac{1}{3} + \frac{(-2)}{3}}]$

$= [(3) * (3)^{-2}]$ ÷ $[(6)^{\frac{1 - 2}{3}}]$

$= [(3)^{1 + (-2)}]$ ÷ $[(6)^{\frac{-1}{3}}]$

$= \frac{[(3)^{-1}]}{[(6)^{\frac{-1}{3}}]}$

$= \frac{[(3)^{-1}]}{[(2 * 3)^{\frac{-1}{3}}]}$

$= \frac{[(3)^{-1}]}{[(2)^{\frac{-1}{3}} * (3)^{\frac{-1}{3}}]}$

$= \frac{2^{\frac{1}{3}} * 3^{\frac{1}{3}}}{3^{1}}$

$=2^{\frac{1}{3}} * 3^{\frac{1}{3} - 1}$

$=2^{\frac{1}{3}} * 3^{\frac{1- 3}{3}}$

$=2^{\frac{1}{3}} * 3^{\frac{-2}{3}}$

$=\frac {2^{\frac{1}{3}}} {3^{\frac{2}{3}}}$

$=(\frac {2} {3^{2}})^{\frac{1}{3}$

$=(\frac {2} {9 })^{\frac{1}{3}$

$= \sqrt[3]{\frac{2}{9} }$

I hope it helps...........