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)

$frac{1}{2x-10} + frac{3}{5x+25} = frac{5}{x^2-25},~~xneq5,~xneq-5\\frac{1}{2x-10} + frac{3}{5x+25} = frac{5}{x^2-25}=0\\frac{1}{2(x-5)} + frac{3}{5(x+5)} = frac{5}{(x-5)(x+5)}\\frac{5(x+5)+6(x-5)-50}{10(x-5)(x+5)} = 0\\frac{11x-55}{10(x-5)(x+5)} = 0\\frac{11(x-5)}{10(x-5)(x+5)}=0\\frac{11}{10(x+5)}=0\\11=0$

$11 neq 0$ - НЕТ КОРНЕЙ

Answer 2

Спасибо большое

Answer 3

Не за что :)