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Question

Bài 1 :Thực hiện phép tính

a, $\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}$

b$\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}$

c, $\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}$

d,...

Nguyễn Lê Phước Thịnh · Accepted Answer

Bài 1:

a) Ta có: $\frac{2x}{x^2+2xy}+\frac{y}{xy-2y^2}+\frac{4}{x^2-4y^2}$

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

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

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

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

b) Ta có: $\frac{1}{x-y}+\frac{3xy}{y^3-x^3}+\frac{x-y}{x^2+xy+y^2}$

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

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

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

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

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

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

c) Ta có: $\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}$

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

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

d) Ta có: $\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}$

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

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

$=\frac{2x^2}{x}=2x$

Bài 2:

a) Ta có: $\frac{4x+1}{2}-\frac{3x+2}{3}$

$=\frac{3\left(4x+1\right)}{6}-\frac{2\left(3x+2\right)}{6}$

$=\frac{12x+3-6x-4}{6}$

$=\frac{6x-1}{6}$

b) Ta có: $\frac{x+3}{x}-\frac{x}{x-3}+\frac{9}{x^2-3x}$

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

$=\frac{x^2-9-x^2+9}{x\left(x-3\right)}=\frac{0}{x\left(x-3\right)}=0$

c) Ta có: $\frac{x+3}{x^2+1}-\frac{1}{x^2+2}$

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

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

$=\frac{x^3+2x^2+2x+5}{\left(x^2+1\right)\left(x^2+2\right)}$

e) Ta có: $\frac{3}{2x^2+2x}+\frac{2x-1}{x^2-1}-\frac{2}{x}$

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

$=\frac{3\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}+\frac{2x\left(2x-1\right)}{2x\left(x+1\right)\left(x-1\right)}-\frac{2\cdot2\cdot\left(x+1\right)\left(x-1\right)}{2x\left(x+1\right)\left(x-1\right)}$

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

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

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

d) Ta có: $\frac{1}{3x-2}-\frac{4}{3x+2}-\frac{-10x+8}{9x^2-4}$

$=\frac{3x+2}{\left(3x-2\right)\left(3x+2\right)}-\frac{4\left(3x-2\right)}{\left(3x+2\right)\left(3x-2\right)}-\frac{-10x+8}{\left(3x-2\right)\left(3x+2\right)}$

$=\frac{3x+2-12x+8+10x-8}{\left(3x-2\right)\left(3x+2\right)}$

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

f) Ta có: $\frac{3x}{5x+5y}-\frac{x}{10x-10y}$

$=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}$

$=\frac{3x\cdot2\cdot\left(x-y\right)}{10\left(x+y\right)\left(x-y\right)}-\frac{x\cdot\left(x+y\right)}{10\left(x-y\right)\left(x+y\right)}$

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

$=\frac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}$Câu trả lời: Bài 1: