Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
a.\(\frac{1-3x}{2}-\frac{x+3}{2}=\frac{1-3x-x-3}{2}=\frac{1-4x-3}{2}=\frac{-4x-2}{2}=\frac{-2\left(2x+1\right)}{2}=-2x-1\)
b. \(\frac{2\left(x+y\right)\left(x-y\right)}{x}-\frac{-2y^2}{x}=\frac{2\left(x^2-y^2\right)+2y^2}{x}=\frac{2x^2-2y^2+2y^2}{x}=2x\)
c. \(\frac{3x+1}{x+y}-\frac{2x-3}{x+y}=\frac{3x+1-2x+3}{x+y}=\frac{x+4}{x+y}\)
d. \(\frac{xy}{2x-y}-\frac{x^2-1}{y-2x}=\frac{xy}{2x-y}-\frac{1-x^2}{2x-y}=\frac{xy-1+x^2}{2x-y}\)
e. \(\frac{4x-1}{3x^2y}-\frac{7x-1}{3x^2y}=\frac{4x-1-7x+1}{3x^2y}=\frac{-3x}{3x^2y}=\frac{-1}{xy}\)
a, \(\frac{x+2y}{8x^2y^5}-\frac{3x^2+2}{12x^4y^4}\)
=\(\frac{\left(x+2y\right)3x^2}{24x^4y^5}-\frac{\left(3x^2+2\right)2y}{24x^4y^5}\)
=\(\frac{3x^3+6x^2y}{24x^4y^5}-\frac{6x^2y+4y}{24x^4y^5}\)
=\(\frac{3x^3+6x^2y-6x^2y-4y}{24x^4y^5}\)
=\(\frac{3x^3-4y}{24x^4y^5}\)
b,\(\frac{y}{xy-5x^2}-\frac{15y-25x}{y^2-25x^2}\)
=\(\frac{y}{x\left(y-5x\right)}-\frac{15y-25x}{\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y\left(y+5x\right)}{x\left(y-5x\right)\left(y+5x\right)}-\frac{\left(15y-25x\right)x}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2+5xy}{x\left(y-5x\right)\left(y+5x\right)}-\frac{15xy-25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2+5xy-15xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y^2-10xy+25x^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{\left(y-5x\right)^2}{x\left(y-5x\right)\left(y+5x\right)}\)
=\(\frac{y-5x}{x\left(y+5x\right)}\)
c,\(\frac{4-x}{x^3+2x}-\frac{x+5}{x^3-x^2+2x-2}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x^3-x^2\right)+\left(2x-2\right)}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{x^2\left(x-1\right)+2\left(x-1\right)}\)
=\(\frac{4-x}{x\left(x^2+2\right)}-\frac{x+5}{\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{\left(4-x\right)\left(x-1\right)}{x\left(x-1\right)\left(x^2+2\right)}-\frac{\left(x+5\right)x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{4x-4-x^2+x}{x\left(x-1\right)\left(x^2+2\right)}-\frac{x^2+5x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{4x-4-x^2+x-x^2-5x}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2x^2-4}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2\left(x^2+2\right)}{x\left(x-1\right)\left(x^2+2\right)}\)
=\(\frac{-2}{x\left(x-1\right)}\)
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: \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)
\(=\frac{3x\cdot2\cdot\left(x-y\right)}{10\left(x+y\right)\left(x-y\right)}-\frac{x\cdot\left(x+y\right)}{10\left(x-y\right)\left(x+y\right)}\)
\(=\frac{6x^2-6xy-x^2-xy}{10\left(x-y\right)\left(x+y\right)}\)
\(=\frac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}\)