Question

Допоможіть будь-ласка якомога швидше ​

Answer 1

1)

$\dfrac{a}{(a-5)^2}\times\dfrac{a+5}{a+5}+\dfrac{a+2}{(5+a)(5-a)}\times\dfrac{a-5}{a-5}=\dfrac{(a^2+5a)+(a^2-3a-10)}{(-a-5)(5-a)^2}=\\\\\dfrac{a^2+5a+a^2-3a-10}{(-a-5)(5-a)^2}=\dfrac{2a^2+2a-10}{(-a-5)(5-a)^2}=\dfrac{2(a^2+a-5)}{(-a-5)(5-a)^2}$

2)

$\dfrac{y+3}{2y+2}\times\dfrac{y-1}{y-1}-\dfrac{y+1}{2y-2}\times\dfrac{y+1}{y+1}+\dfrac{3}{y^2-1}\times\dfrac{2}{2}=\dfrac{(y^2+2y-3)-(y^2+2y+1)+6}{2y^2-2}=\\\dfrac{(y^2+2y-3)-(y^2+2y+1)+6}{2y^2-2}=\dfrac{2}{2y^2-2}=\dfrac{1}{y^2-1}$

3)

$\dfrac{2b^2-b}{(b+1)(b^2-b+1)}\times\dfrac{b+1}{b+1}-\dfrac{b-1}{b^2-b+1}\times\dfrac{(b+1)^2}{(b+1)^2}=\dfrac{(2b^3+b^2-b)-(b^3+b^2-b-1)}{(b+1)^2(b^2-b+1)}=\\\dfrac{b^3+1}{(b+1)^2(b^2-b+1)}=\dfrac{b^3+1}{(b^3+1)(b+1)}=\dfrac{1}{b+1}$