Question

МАТЕМАТИКА 9 КЛАСС (2.В.36 а и б)

Answer 1

$a); ; frac{frac{1}{sqrt{a-1}}+sqrt{a+1}}{frac{1}{sqrt{a+1}}-frac{1}{sqrt{a-1}}}: frac{sqrt{a+1}}{(a-1)sqrt{a+1}-(a+1)sqrt{a-1}}=\\=frac{(1+ sqrt{(a-1)(a+1)})cdot sqrt{(a+1)(a-1)}}{sqrt{a-1}cdot (sqrt{a-1}-sqrt{a+1})}cdot frac{sqrt{(a-1)(a+1)}cdot (sqrt{a-1}-sqrt{a+1})}{sqrt{a+1}}=\\=frac{(1+sqrt{a^2-1})cdot (sqrt{(a+1)(a-1)})^2}{sqrt{(a-1)(a+1)}}=(1+sqrt{a^2-1})cdot sqrt{(a+1)(a-1)}=\\=(1+sqrt{a^2-1})cdot sqrt{a^2-1}=sqrt{a^2-1}+a^2-1$

$b); ; frac{frac{1}{sqrt{a-1}}+sqrt{a-1}}{frac{1}{sqrt{a+1}}-frac{1}{sqrt{a-1}}}:frac{sqrt{a-1}}{(a-1)sqrt{a+1}-(a+1)sqrt{a-1}}=\\=frac{(1+(a-1))sqrt{(a-1)(a+1)}}{sqrt{a-1}(sqrt{a-1}-sqrt{a+1})}cdot frac{sqrt{(a-1)(a+1)}(sqrt{a-1}-sqrt{a+1})}{sqrt{a-1}}=\\=frac{acdot (sqrt{(a-1)(a+1)})^2}{(sqrt{a-1})^2}=frac{acdot (a-1)(a+1)}{a-1}=a(a+1)$

Answer 2

Огроооомное спасибоо))