Question

помагите пожалуйста ​

Answer 1

$1)\frac{a^{2}+1 }{a-3}-\frac{10}{a-3} =\frac{a^{2}+1-10 }{a-3}=\frac{a^{2}-9 }{a-3}=\frac{(a-3)(a+3)}{a-3}=a+3\\\\a=97 \ \Rightarrow \ a+3=97+3=\boxed{100}\\\\\\2)\frac{x+7}{x^{2}-25 }-\frac{2x+2}{x^{2}-25 } =\frac{x+7-2x-2}{x^{2}-25 }=\frac{5-x}{(x-5)(x+5)}=-(x+5)\\\\x=-5,1 \ \Rightarrow \ -(x+5)=-(-5,1+5)=-(-0,1)=\boxed{0,1}$

6,47)

$1)\frac{3}{x+y}-\frac{5}{x}=\frac{3x-5(x+y)}{x(x+y)}=\frac{3x-5x-5y}{x(x+y)}=\frac{-2x-5y}{x(x+y)} =-\frac{2x+5y}{x(x+y)} \\\\\\2)\frac{4}{a-b}+\frac{1}{a}=\frac{4a+a-b}{a(a-b)}=\frac{5a-b}{a(a-b)}\\\\\\3)\frac{6}{m-1}-\frac{2}{m} =\frac{6m-2(m-1)}{m(m-1)}=\frac{6m-2m+2}{m(m-1)}=\frac{4m+2}{m(m-1)}\\\\\\4)\frac{1}{b+2} -\frac{3}{b} =\frac{b-3(b+2)}{b(b+2)}=\frac{b-3b-6}{b(b+2)}=-\frac{2b+6}{b(b+2)}$