Question

plz ans it..........

Answer 1

20. and. 21..........

Answer 2

Answer: 20) -4√6

21)  7-3√5

Step-by-step explanation:

20) Since, $a^2+b^2-5ab = (a-b)^2-3ab$

Here,

$a=\frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}$

$b=\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$\implies ab = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}}\times \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$\implies ab = \frac{1}{1}=1$

Now,

$(a-b) = \frac{\sqrt{3}-\sqrt{2}}{\sqrt{3}+\sqrt{2}} - \frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}-\sqrt{2}}$

$=\frac{(\sqrt{3}-\sqrt{2})^2-(\sqrt{3}+\sqrt{2})^2}{3-2}$

$=\frac{-4\sqrt{6}}{1}=-4\sqrt{6}$

$\implies (a-b)^2=96$

$a^2+b^2-5ab = (a-b)^2-3ab=96-3 = 93$

21) $p^2+q^2= (p+q)^2-2pq$

Here,

$p=\frac{3-\sqrt{5}}{3+\sqrt{5}}$

$q=\frac{3+\sqrt{5}}{3-\sqrt{5}}$

$\implies p+q=\frac{3-\sqrt{5}}{3+\sqrt{5}}+\frac{3+\sqrt{5}}{3-\sqrt{5}}$

$=\frac{(3-\sqrt{5})^2+(3+\sqrt{5})^2}{9-5}$

$=\frac{18+10}{4}$

$=\frac{28}{4}=7$

Also,

$pq = \frac{3-\sqrt{5}}{3+\sqrt{5}}\times \frac{3+\sqrt{5}}{3-\sqrt{5}}=1$

Hence,

$p^2+q^2= (p+q)^2-2pq=49 - 2 = 47$