Dạng 9: Bài luyện tập có đáp án

$\frac{2x-1}{4x^{2}y}-\frac{6x-1}{4x^{2}y}=\frac{8x}{4x^{2}y}=\frac{2}{xy}$

$\frac{3x+2}{2x-3}-\frac{2-5x}{2x-3}=\frac{8x}{2x-3}$

$\frac{x-17}{3-2x}-\frac{11x+1}{2x-3}=\frac{x-17}{3-2x}+\frac{11x+1}{3-2x}=\frac{12x-18}{-2x+3}=-6$

$\frac{10x-15}{3x-2}-\frac{2x+7}{2-3x}=\frac{10x-15}{3x-2}+\frac{2x+7}{3x-2}=\frac{12x-8}{3x-2}=4$

$\frac{2xy-5}{8x{y}^2}-\frac{5x{y}^2-5}{8x{y}^2}=\frac{2xy-5x{y}^2}{8x{y}^2}=\frac{2-5y}{8y}$

$\frac{7x+5}{3x(x+4)}-\frac{4x+5}{3x^2+12x}=\frac{7x+5}{3x(x+4)}-\frac{4x+5}{3x(x+4)}=\frac{3x}{3x(x+4)}=\frac{1}{x+4}$

$\frac{4}{3x+2}-\frac{2x+1}{3x^2+2x}=\frac{4}{3x+2}-\frac{2x+1}{x(3x+2)}=\frac{2x-1}{x(3x+2)}$

$x^2+2-\frac{x^4-2x^2+2}{x^2-2}=\frac{x^4-4-x^4+2x^2-2}{x^2-2}=\frac{2x^2-6}{x^2-2}$

$\frac{x-1+1-y}{x-y}=\frac{x-y}{x-y}=1$

$\frac{x^2-1-\left(x^2-4\right)}{x+1}=\frac{3}{x+1}$

$\frac{1}{x-1}-\frac{2x}{\left(x-1\right)\left(x^2+1\right)}=\frac{x^2+1-2x}{\left(x-1\right)\left(x^2+1\right)}=\frac{x-1}{x^2+1}$

$\frac{3x-1}{2\left(3x+1\right)}+\frac{3x+1}{2\left(3x-1\right)}-\frac{6x}{9x^2-1}=\frac{{\left(3x-1\right)}^2+{\left(3x+1\right)}^2-12x}{2\left(3x+1\right)\left(3x-1\right)}=\frac{18x^2-12x+2}{2\left(3x+1\right)\left(3x-1\right)}$

$=\frac{2{\left(3x-1\right)}^2}{2\left(3x+1\right)\left(3x-1\right)}=\frac{3x-1}{3x+1}$

=0

$\frac{x+1}{x^3+1}=\frac{1}{x^2-x+1}$

$\begin{array}{l}=\frac{(x+2)(x+3)-2(x+1)(x+3)+(x+1)(x+2)}{(x+1){(x+2)}^2(x+3)}\\ =\frac{x^2+5x+6-2x^2-8x-6+x^2+3x+2}{(x+1){(x+2)}^2(x+3)}\\ =\frac{2}{(x+1){(x+2)}^2(x+3)}\end{array}$

$\frac{2x^2-7x-12}{(x+1)(2x+3)(2x-3)}$

$\begin{array}{l}=\frac{x^2-x-2+x^2+x-2+x^3-3x-2}{(x-2)(x-1)(x+1)(x+2)}\\ =\frac{x^3+2x^2-3x-6}{(x-2)(x-1)(x+1)(x+2)}\\ =\frac{x^2-3}{(x-2)(x-1)(x+1)}\end{array}$

$y-2x=5\iff y=2x+5$

$\frac{3x-y}{x-5}-\frac{2x-3y}{2y+5}=\frac{3x-y}{x-5}+\frac{3y-2x}{2y+5}=\frac{x-\left(y-2x\right)}{x-5}+\frac{2y+\left(y-2x\right)}{2y+5}$

$=\frac{x-5}{x-5}+\frac{2y+5}{2y+5}=1+1=2$

$\frac{4x^2-{(x-3)}^2}{9(x^2-1)}-\frac{x^2-9}{{(2x+3)}^2-x^2}+\frac{{(2x-3)}^2-x^2}{4x^2-{(x+3)}^2}$

$=\frac{(2x-x+3)(2x+x-3)}{9(x-1)(x+1)}-\frac{(x-3)(x+3)}{(2x+3-x)(2x+3+x)}+\frac{(2x-3-x)(2x-3+x)}{(2x-x-3)(2x+x+3)}$

$=\frac{3(x+3)(x-1)}{9(x-1)(x+1)}-\frac{(x-3)(x+3)}{3(x+3)(x+1)}+\frac{3(x-3)(x-1)}{3(x-3)(x+1)}$

$=\frac{x+3}{3(x+1)}-\frac{x-3}{3(x+1)}+\frac{3(x-1)}{3(x+1)}=\frac{x+3-x+3+3x-3}{3(x+1)}=\frac{3x+3}{3x+3}=1$

$\frac{y-z}{(x-y)(x-z)}+\frac{z-x}{(y-z)(y-x)}+\frac{x-y}{(z-x)(z-y)}$

$=\frac{(x-z)-(x-y)}{(x-y)(x-z)}+\frac{(y-x)-(y-z)}{(y-z)(y-x)}+\frac{(z-y)-(z-x)}{(z-x)(z-y)}$

$=\frac{1}{x-y}-\frac{1}{x-z}+\frac{1}{y-z}-\frac{1}{y-x}+\frac{1}{z-x}-\frac{1}{z-y}$

$=\frac{1}{x-y}+\frac{1}{z-x}+\frac{1}{y-z}+\frac{1}{x-y}+\frac{1}{z-x}+\frac{1}{y-z}$

$=\frac{2}{x-y}+\frac{2}{y-z}+\frac{2}{z-x}$

$\begin{array}{l}\frac{1}{x(x+1)}+\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}\\ =\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+4}\\ =\frac{1}{x}-\frac{1}{x+4}\\ =\frac{x+3}{x(x+4)}\end{array}$

$\begin{array}{l}\frac{1}{x(x+1)}+\frac{1}{(x+1)(x+2)}+\frac{1}{(x+2)(x+3)}+\frac{1}{(x+3)(x+4)}+\frac{1}{(x+4)(x+5)}+\frac{1}{x+5}\\ =\frac{1}{x}-\frac{1}{x+5}+\frac{1}{x+5}=\frac{1}{x}\end{array}$

$\frac{1550}{x-1}-\frac{1500}{x}=\frac{1550x-1500x+1500}{x}=\frac{50x+1500}{x}$