Đáp án:
$a)-\dfrac{2x+y}{xy}\\ b)\dfrac{2}{x+7}\\ c)\dfrac{1}{x}\\ d)\dfrac{1−5x}{x(5x+1)}\\ e)\dfrac{6x}{x-3}\\ f)\dfrac{8}{2x+3}$
Giải thích các bước giải:
$a)\dfrac{y}{2x^2-xy}+\dfrac{4x}{y^2-2xy}\\ =\dfrac{y}{x(2x-y)}+\dfrac{4x}{y(y-2x)}\\ =\dfrac{y}{x(2x-y)}-\dfrac{4x}{y(2x-y)}\\ =\dfrac{y^2}{xy(2x-y)}-\dfrac{4x^2}{xy(2x-y)}\\ =\dfrac{y^2-4x^2}{xy(2x-y)}\\ =\dfrac{(y-2x)(y+2x)}{xy(2x-y)}\\ =-\dfrac{2x+y}{xy}\\ b)\dfrac{7x+6}{2x(x+7)}-\dfrac{3x+6}{2x^2+14x}\\ =\dfrac{7x+6}{2x(x+7)}-\dfrac{3x+6}{2x(x+7)}\\ =\dfrac{7x+6-(3x+6)}{2x(x+7)}\\ =\dfrac{4x}{2x(x+7)}\\ =\dfrac{2}{x+7}\\ c)\dfrac{4x+13}{5x(x-7)}-\dfrac{x-48}{35x-5x^2}\\ =\dfrac{4x+13}{5x(x-7)}+\dfrac{x-48}{5x^2-35x}\\ =\dfrac{4x+13}{5x(x-7)}+\dfrac{x-48}{5x(x-7)}\\ =\dfrac{4x+13+x-48}{5x(x-7)}\\ =\dfrac{5x-35}{5x(x-7)}\\ =\dfrac{5(x-7)}{5x(x-7)}\\ =\dfrac{1}{x}\\ d)\dfrac{1}{x-5x^2}-\dfrac{25x-15}{25x^2-1}\\ =\dfrac{1}{x(1-5x)}-\dfrac{25x-15}{(5x-1)(5x+1)}\\ =-\dfrac{5x+1}{x(5x-1)(5x+1)}-\dfrac{(25x-15)x}{x(5x-1)(5x+1)}\\ =\dfrac{-(5x+1)-(25x-15)x}{x(5x-1)(5x+1)}\\ =\dfrac{−25x^2+10x−1}{x(5x-1)(5x+1)}\\ =\dfrac{−(5x-1)^2}{x(5x-1)(5x+1)}\\ =\dfrac{−(5x-1)}{x(5x+1)}\\ =\dfrac{1−5x}{x(5x+1)}\\ e)\dfrac{6x}{x^2-9}-\dfrac{5x}{3-x}+\dfrac{x}{x+3}\\ =\dfrac{6x}{(x-3)(x+3)}+\dfrac{5x}{x-3}+\dfrac{x}{x+3}\\ =\dfrac{6x}{(x-3)(x+3)}+\dfrac{5x(x+3)}{(x-3)(x+3)}+\dfrac{x(x-3)}{(x-3)(x+3)}\\ =\dfrac{6x+5x(x+3)+x(x-3)}{(x-3)(x+3)}\\ =\dfrac{6x^2+18x}{(x-3)(x+3)}\\ =\dfrac{6x(x+3)}{(x-3)(x+3)}\\ =\dfrac{6x}{x-3}\\ f)\dfrac{5x}{2x^2-3x}+\dfrac{2}{2x+3}-\dfrac{2x-33}{9-4x^2}\\ =\dfrac{5x}{2x^2-3x}+\dfrac{2}{2x+3}+\dfrac{2x-33}{4x^2-9}\\ =\dfrac{5x}{x(2x-3)}+\dfrac{2}{2x+3}+\dfrac{2x-33}{(2x-3)(2x+3)}\\ =\dfrac{5x(2x+3)}{x(2x-3)}+\dfrac{2x(2x-3)}{x(2x-3)(2x+3)}+\dfrac{(2x-33)x}{x(2x-3)(2x+3)}\\ =\dfrac{5x(2x+3)+2x(2x-3)+(2x-33)x}{x(2x-3)(2x+3)}\\ =\dfrac{16x^2−24x}{x(2x-3)(2x+3)}\\ =\dfrac{8x(2x−3)}{x(2x-3)(2x+3)}\\ =\dfrac{8}{2x+3}$
