Question

я знаю что тут есть умные и добрые люди.

упростить выражение ​

Answer 1

Решение задания прилагаю

Answer 2

Решение:

$4)~\dfrac{3x+2}{x^2 - 2x + 1} - \dfrac{6}{x^2 - 1} - \dfrac{3x-2}{x^2 + 2x + 1} =\\ \\ =\dfrac{3x+2}{(x-1)^2} - \dfrac{6}{(x - 1)(x +1)} - \dfrac{3x-2}{(x+1)^2} =\\ \\ = \dfrac{(3x+2)(x+1)^2}{(x-1)^2(x+1)^2} - \dfrac{6(x+1)(x-1)}{(x - 1)^2(x +1)^2} - \dfrac{(3x-2)(x-1)^2}{(x+1)^2(x-1)^2} =\\ \\ = \dfrac{3x((x +1)^2- (x-1)^2)+2((x+1)^2 + (x - 1)^2)-6(x^2 -1)}{(x -1)^2(x+1)^2} =\\ \\ =\dfrac{3x(x^2 +2x + 1 -x^2 +2x-1)}{(x+1)^2(x-1)^2} +\dfrac{2(x^2 +2x +1 +x^2 -2x +1)}{(x+1)^2(x-1)^2} -$

$-\dfrac{6(x^2 -1)}{y} = \dfrac{3x\cdot 4x +4x^2+4-6x^2 + 6}{(x-1)^2(x+1)^2} =\dfrac{10(x^2 + 1)}{(x-1)^2(x+1)^2} = \dfrac{10(x^2 + 1)}{(x-1)^2(x+1)^2} =\dfrac{10(x^2 + 1)}{(x^2-1)^2}$

$5)~ \dfrac{2a^2 + 7}{a^2 + 3a+9} -\dfrac{a}{3-a} -\dfrac{3a^3-2a^2+19a - 12}{a^3 -27} =\\ \\ =\dfrac{(2a^2 + 7)(a-3) +a(a^2 + 3a+9) - (3a^3-2a^2+19a - 12)}{a^3 -27} =\\ \\ =\dfrac{2a^3 +7a -6a^2-21 +a^3 + 3a^2 +9a -3a^3+2a^2-19a+12}{a^3-27} =\\ \\ =\dfrac{-a^2-3a-9}{a^3-27}=-\dfrac{(a^2+3a +9)}{a^3-27} = \dfrac{1}{3-a}$