a)\(\frac{1}{x-y}-\frac{3xy}{x^3-y^3}+\frac{x+y}{x^2+xy+y^2}\)
b)\(\frac{x+2}{x^2}+\frac{3}{2x}+\frac{-5x}{2x^2}\)
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KN
29 tháng 11 2019
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
KN
29 tháng 11 2019
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
19 tháng 8 2020
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: \(\frac{3x}{5x+5y}-\frac{x}{10x-10y}\)
\(=\frac{3x}{5\left(x+y\right)}-\frac{x}{10\left(x-y\right)}\)
\(=\frac{3x\cdot2\cdot\left(x-y\right)}{10\left(x+y\right)\left(x-y\right)}-\frac{x\cdot\left(x+y\right)}{10\left(x-y\right)\left(x+y\right)}\)
\(=\frac{6x^2-6xy-x^2-xy}{10\left(x-y\right)\left(x+y\right)}\)
\(=\frac{5x^2-7xy}{10\left(x-y\right)\left(x+y\right)}\)
KN
20 tháng 11 2019
a) \(\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}:\frac{10x-10y}{x^3+y^3}\)
\(=\frac{3x^2-6xy+3y^2}{5x^2-5xy+5y^2}.\frac{x^3+y^3}{10x-10y}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5\left(x^2-xy+y^2\right)}.\frac{\left(x+y\right)\left(x^2-xy+y^2\right)}{10\left(x-y\right)}\)
\(=\frac{3\left(x^2-2xy+y^2\right)}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)^2}{5}.\frac{x+y}{10\left(x-y\right)}\)
\(=\frac{3\left(x-y\right)}{5}.\frac{x+y}{10}\)
\(=\frac{3x^2-3y^2}{50}\)
KN
20 tháng 11 2019
c) \(\frac{2}{xy}:\left(\frac{1}{x}-\frac{1}{y}\right)-\frac{x^2-y^2}{\left(x-y\right)^2}\)
\(=\frac{2}{xy}:\frac{y-x}{xy}-\frac{\left(x+y\right)\left(x-y\right)}{\left(x-y\right)^2}\)
\(=\frac{2}{y-x}-\frac{x+y}{x-y}\)
\(=\frac{2}{y-x}+\frac{x+y}{y-x}\)
\(=\frac{x+y+2}{y-x}\)
KN
7 tháng 11 2019
a) Ta có: \(\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)\)
\(=x\left(x^4+x^3+x^2+x+1\right)\)\(-\left(x^4+x^3+x^2+x+1\right)\)
\(=x^5+x^4+x^3+x^2+x-x^4-x^3-x^2-x-1\)
\(=x^5-1\)
Vậy \(\left(x-1\right)\left(x^4+x^3+x^2+x+1\right)\)\(=x^5-1\)
hay \(\frac{x^5-1}{x-1}=x^4+x^3+x^2+x+1\)
VM
1 tháng 2 2020
Bài 4:
a) \(\frac{2x^2-10xy}{2xy}+\frac{5y-x}{y}\)
\(=\frac{y.\left(2x^2-10xy\right)}{2xy.y}+\frac{2xy.\left(5y-x\right)}{2xy.y}\)
\(=\frac{2x^2y-10xy^2}{2xy^2}+\frac{10xy^2-2x^2y}{2xy^2}\)
\(=\frac{2x^2y-10xy^2+10xy^2-2x^2y}{2xy^2}\)
\(=\frac{0}{2xy^2}\)
\(=0.\)
b) \(\frac{2}{x+y}+\frac{1}{x-y}+\frac{3x}{x^2-y^2}\)
\(=\frac{2}{x+y}+\frac{1}{x-y}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2.\left(x-y\right)}{\left(x-y\right).\left(x+y\right)}+\frac{1.\left(x+y\right)}{\left(x-y\right).\left(x+y\right)}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2x-2y}{\left(x-y\right).\left(x+y\right)}+\frac{x+y}{\left(x-y\right).\left(x+y\right)}+\frac{3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{2x-2y+x+y+3x}{\left(x-y\right).\left(x+y\right)}\)
\(=\frac{6x-y}{\left(x-y\right).\left(x+y\right)}\)
c) \(x+y+\frac{x^2+y^2}{x+y}\)
\(=\frac{x+y}{1}+\frac{x^2+y^2}{x+y}\)
\(=\frac{\left(x+y\right).\left(x+y\right)}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{\left(x+y\right)^2}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{x^2+2xy+y^2}{x+y}+\frac{x^2+y^2}{x+y}\)
\(=\frac{x^2+2xy+y^2+x^2+y^2}{x+y}\)
\(=\frac{2x^2+2xy+2y^2}{x+y}.\)
Chúc bạn học tốt!