Question

Помогите,пожалуйста, сделать задания №3 и №4

Answer 1

3)

$\frac{b}{a-b} -\frac{a^2-b^2}{a+3b} * (\frac{a+b}{(a-b)^2} + \frac{b}{a^2-b^2}) = - 1\\\\\frac{b}{a-b} - \frac{(a-b)(a+b)}{a+3b} * \frac{(a+b)^2+b(a-b)}{(a-b)^2*(a+b)} = -1 \\\\\frac{b}{a-b} - \frac{1}{a+3b} * \frac{(a+b)^2+ab-b^2}{a-b} = -1\\\\\frac{b}{a-b} - \frac{(a+b)^2+ab-b^2}{(a+3b)(a-b)} = -1\\\\\frac{b}{a-b} - \frac{a^2+2ab+b^2+ab-b^2}{(a+3b)(a-b)} = -1\\\\\frac{b}{a-b} - \frac{a^2+3ab}{(a+3b)(a-b)} = -1\\\\\frac{b}{a-b} - \frac{a(a+3b)}{(a+3b)(a-b)} = -1$

$\frac{b}{a-b} -\frac{a}{a-b} = -1\\\\\frac{b-a}{a-b} = -1\\\\\frac{-(a-b)}{a-b} = -1 \\\\-1 = -1$

4)

$\frac{1}{2x-10} + \frac{3}{5x+25} = \frac{5}{x^2-25},~~x\neq5,~x\neq-5\\\\\frac{1}{2x-10} + \frac{3}{5x+25} = \frac{5}{x^2-25}=0\\\\\frac{1}{2(x-5)} + \frac{3}{5(x+5)} = \frac{5}{(x-5)(x+5)}\\\\\frac{5(x+5)+6(x-5)-50}{10(x-5)(x+5)} = 0\\\\\frac{11x-55}{10(x-5)(x+5)} = 0\\\\\frac{11(x-5)}{10(x-5)(x+5)}=0\\\\\frac{11}{10(x+5)}=0\\\\11=0$

$11 \neq 0$ - НЕТ КОРНЕЙ