Dạng 1: bài luyện tập số 1 có đáp án

$x-2-\frac{x^2-10}{x+2}=\frac{\left(x-2\right)\left(x+2\right)-x^2+10}{x+2}=\frac{6}{x+2}$

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

$=\frac{2x+6}{2x\left(x+3\right)}=\frac{2\left(x+3\right)}{2x\left(x+3\right)}=\frac{1}{x}$

$\frac{x-3}{4x+4}-\frac{x-1}{6x-30}$$=\frac{x-3}{4\left(x+1\right)}+\frac{x-1}{6\left(5-x\right)}=\frac{3\left(x-3\right)\left(5-x\right)+2\left(x-1\right)\left(x+1)\right)}{12\left(x+1\right)\left(5-x\right)}$$=\frac{-5x^2+2x-47}{12\left(x+1\right)\left(5-x\right)}$

$\frac{1}{x-5x^2}-\frac{25x-15}{25x^2-1}$$=\frac{1}{x\left(1-5x\right)}+\frac{25x-15}{\left(1-5x\right)\left(1+5x\right)}=\frac{1+5x+x\left(25x-15\right)}{x\left(1-5x\right)\left(1+5x\right)}$$=\frac{1+25x^2-10x}{x\left(1-5x\right)\left(1+5x\right)}=\frac{{(1-5x)}^2}{x\left(1-5x\right)\left(1+5x\right)}=\frac{1-5x}{x\left(1+5x\right)}$

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

$\begin{array}{l}=\frac{x^4+y^4+2x^2{y}^2-2x^4-2y^4}{x^2+y^2}\\ =\frac{-{\left(x^2-y^2\right)}^2}{x^2+y^2}\end{array}$

$\frac{x+9y}{x^2-9y^2}-\frac{3y}{x^2+3xy}$$\frac{x\left(x+9y\right)-3y\left(x-3y\right)}{x\left(x-3y\right)\left(x+3y\right)}=\frac{x^2+6xy+9y^2}{x\left(x-3y\right)\left(x+3y\right)}$

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

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

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

$\frac{3y+4}{5y-10}-\frac{y+4}{3y-6}=\frac{3y+4}{5(y-2)}-\frac{y+4}{3\left(y-2\right)}=\frac{9y+12-5y-20}{15(y-2)}$$=\frac{4y-8}{15(y-2)}=\frac{4(y-2)}{15(y-2)}=\frac{4}{15}$

Ta có:$\frac{1}{x}-\frac{1}{x+3}=\frac{x+3}{x\left(x+3\right)}-\frac{x}{x\left(x+3\right)}=\frac{3}{x\left(x+3\right)}\Rightarrow$  ĐPCM.

 Áp dụng, ta có:

$3M=\frac{3}{x\left(x+3\right)}+\frac{3}{\left(x+3\right)\left(x+6\right)}+\mathrm{...}+\frac{3}{\left(x+12\right)\left(x+15\right)}$         

 = $\frac{1}{x}-\frac{1}{x+3}+\frac{1}{x+3}-\frac{1}{x+6}+\mathrm{...}+\frac{1}{x+12}-\frac{1}{x+15}$

$=\frac{1}{x}-\frac{1}{x+15}=\frac{15}{x\left(x+15\right)}\Rightarrow$$M=\frac{5}{x\left(x+15\right)}.$

$\frac{1}{x(x+1)}+\frac{1}{\left(x+1\right)(x+2)}+\frac{1}{\left(x+2\right)(x+3)}+\mathrm{........}+\frac{1}{\left(x+2007\right)(x+2008)}$

$=\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+3}+\mathrm{.........}+\frac{1}{x+2007}-\frac{1}{x+2008}$

$=\frac{1}{x}-\frac{1}{x+2008}=\frac{2008}{x(x+2008)}$

$\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)}$ư

$=\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$

$M=\frac{a^4-16}{a^4-4a^3+8a^2-16a+16}=\frac{\left(a^2+2^2\right)\left(a^2-2^2\right)}{{\left(a-2\right)}^2\left(a^2+4\right)}$$=\frac{\left(a^2+2^2\right)\left(a-2\right)\left(a+2\right)}{{\left(a-2\right)}^2\left(a^2+4\right)}=\frac{a+2}{a-2}$

$\frac{10x-4}{x^3-4x}=\frac{a}{x}+\frac{b}{x+2}+\frac{c}{x-2}$

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

$-\frac{x+1}{5-x}=\frac{x+1}{-\left(5-x\right)}=\frac{x+1}{x-5}$

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

$\frac{7x+3}{2x^{2}y}-\frac{3+x}{2x^{2}y}=\frac{7x+3-3-x}{2x^{2}y}=\frac{8x}{2x^{2}y}=\frac{4}{xy}$

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

$\frac{3x+2}{2x-5}-\frac{7x+3}{2x-5}=\frac{3x+2-7x-3}{2x-5}=\frac{-4x-1}{2x-5}$

$\frac{2x+3}{x-5}-\frac{x+7}{5-x}=\frac{2x+3}{x-5}+\frac{x+7}{x-5}=\frac{2x+3+x+7}{x-5}=\frac{3x+10}{x-5}$

$b)\frac{4x-15}{3x(x-3)}-\frac{2x-3}{3x(3-x)}=\frac{4x-15}{3x(x-3)}+\frac{2x-3}{3x(x-3)}=\frac{4x-15+2x-3}{3x(x-3)}=\frac{6x-18}{3x(x-3)}=\frac{6(x-3)}{3x(x-3)}=\frac{2}{x}$

$a)\frac{x+5y}{x{y}^2}-\frac{5x-3y}{x^{2}y}=\frac{x(x+5y)}{x^2{y}^2}-\frac{y(5x-3y)}{x^2{y}^2}=\frac{x(x+5y)-y(5x-3y)}{x^2{y}^2}=\frac{x^2+5xy-5xy+3y^2}{x^2{y}^2}=\frac{x^2+3y^2}{x^2{y}^2}$

$a)\frac{x+5y}{x{y}^2}-\frac{5x-3y}{x^{2}y}=\frac{x(x+5y)}{x^2{y}^2}-\frac{y(5x-3y)}{x^2{y}^2}=\frac{x(x+5y)-y(5x-3y)}{x^2{y}^2}=\frac{x^2+5xy-5xy+3y^2}{x^2{y}^2}=\frac{x^2+3y^2}{x^2{y}^2}$

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

$\begin{array}{l}A+\frac{3}{x-2}=\frac{4x+5}{x^2-4}\\ A=\frac{4x+5}{x^2-4}-\frac{3}{x-2}\\ A=\frac{4x+5}{(x-2)(x+2)}-\frac{3(x+2)}{(x-2)(x+2)}\\ A=\frac{4x+5-3x-6}{(x-2)(x+2)}\\ A=\frac{x-1}{(x-2)(x+2)}\end{array}$

$\begin{array}{l}A+\frac{2x-5}{3x+4}=\frac{2x^2+4x-5}{3x^2+4x}\\ A=\frac{2x^2+4x-5}{3x^2+4x}-\frac{2x-5}{3x+4}\\ A=\frac{2x^2+4x-5}{x(3x+4)}-\frac{(2x-5)x}{(3x+4)x}\\ A=\frac{2x^2+4x-5-2x^2+5x}{x(3x+4)}\\ A=\frac{9x-5}{x(3x+4)}\end{array}$

$\begin{array}{l}\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}\\ =\frac{3x+2}{(3x-2)(3x+2)}-\frac{4(3x-2)}{(3x-2)(3x+2)}-\frac{-10x+8}{(3x-2)(3x+2)}\\ =\frac{3x+2-12x+8+10x-8}{(3x-2)(3x+2)}\\ =\frac{x+2}{(3x-2)(3x+2)}\end{array}$

$\begin{array}{l}\frac{x+3}{x}-\frac{x}{x-3}-\frac{9}{3x-x^2}\\ =\frac{(x-3)(x+3)}{x(x-3)}-\frac{x.x}{x(x-3)}+\frac{9}{x(x-3)}\\ =\frac{x^2-9-x^2+9}{x(x-3)}=0\end{array}$

$\begin{array}{l}\frac{4x^2-3x+5}{x^3-1}-\frac{1-2x}{x^2+x+1}-\frac{6}{x-1}\\ =\frac{4x^2-3x+5}{(x-1)(x^2+x+1)}-\frac{(1-2x)(x-1)}{(x^2+x+1)(x-1)}-\frac{6(x^2+x+1)}{(x-1)(x^2+x+1)}\\ =\frac{4x^2-3x+5+2x^2-3x+1-6x^2-6x-6}{(x-1)(x^2+x+1)}\\ =\frac{-12x}{(x-1)(x^2+x+1)}\end{array}$

$\begin{array}{l}\frac{3}{2x^2+2x}-\frac{1-2x}{x^2-1}-\frac{2}{x}\\ =\frac{3(x-1)}{2x(x+1)(x-1)}-\frac{2x(1-2x)}{2x(x+1)(x-1)}-\frac{2.2(x-1)(x+1)}{2x(x+1)(x-1)}\\ =\frac{3x-3-2x+4x^2-4x^2+4}{2x(x-2)(x+2)}\\ =\frac{x+1}{2x(x-2)(x+2)}\end{array}$

$\begin{array}{l}A=\frac{1}{x(x+1)}+\frac{1}{(x+1)(x+2)}+\mathrm{...}+\frac{1}{(x+2019)(x+2020)}\\ =\frac{1}{x}-\frac{1}{x+1}+\frac{1}{x+1}-\frac{1}{x+2}+\mathrm{....}+\frac{1}{x+2019}-\frac{1}{x+2020}\\ =\frac{1}{x}-\frac{1}{x+2020}\\ =\frac{x+2020}{x(x+2020)}-\frac{x}{x(x+2020)}\\ =\frac{2020}{x(x+2020)}\end{array}$$Vậy$$A=\frac{2020}{x(x+2020)}$

$\begin{array}{l}B=\frac{1}{x(x+2)}+\frac{1}{(x+2)(x+4)}+\frac{1}{(x+4)(x+6)}+\mathrm{...}+\frac{1}{(x+2018)(x+2020)}\\ 2B=\frac{2}{x(x+2)}+\frac{2}{(x+2)(x+4)}+\frac{2}{(x+4)(x+6)}+\mathrm{...}+\frac{2}{(x+2018)(x+2020)}\\ 2B=\frac{1}{x}-\frac{1}{x+2}+\frac{1}{x+2}-\frac{1}{x+4}+\mathrm{...}+\frac{1}{x+2018}-\frac{1}{x+2020}\\ 2B=\frac{1}{x}-\frac{1}{x+2020}\\ 2B=\frac{2020}{x(x+2020)}\\ B=\frac{1010}{x(x+2020)}\end{array}$