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a) \(\frac{x^2-xy}{5y^2-5xy}\)=\(\frac{x\left(x-y\right)}{-5y\left(x-y\right)}\)=\(\frac{-x}{5y}\)
b) \(\frac{3x^2-12x+12}{x^4-8x}\)=\(\frac{3\left(x^2-4x+4\right)}{x\left(x^3-2^3\right)}\)=\(\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)=\(\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)
\(=\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)\left(x+y\right)^2}\)
\(=\frac{10y}{15\left(x+y\right)^2}\)
\(\frac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\frac{\left(x^2-x\right)-\left(xy-y\right)}{\left(x^2-x\right)+\left(xy-y\right)}\)
\(=\frac{x\left(x-1\right)-y\left(x-1\right)}{x\left(x-1\right)+y\left(x-1\right)}\)
\(=\frac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)
\(=\frac{x-y}{x+y}\)
a)\(\frac{2xy}{3\left(x+y\right)^2}\)
b)=\(\frac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)=\(\frac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)
=\(\frac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)=\(\frac{\left(x-y\right)}{\left(x+y\right)}\)
a)\(\frac{3xy+6}{6xy+12}=\frac{1}{2}\Leftrightarrow\left(3xy+6\right)\cdot2=\left(6xy+12\right)\cdot1\)
\(\Leftrightarrow6xy+12=6xy+12\)
Vậy.......
b)\(\frac{x^2-xy}{5y^2-5xy}=\frac{x}{5y}\Leftrightarrow\left(x^2-xy\right)\cdot5y=\left(5y^2-5xy\right)\cdot x\)
\(\Leftrightarrow5x^2y-5xy^2=5xy^2-5x^2y\)
Vậy.....
\(\frac{x^8-1}{\left(x^4+1\right)\left(x^2-1\right)}\)
\(=\frac{\left(x^2-1\right)\left(x^4+x^2+1\right)}{\left(x^4+1\right)\left(x^2-1\right)}\)
\(=\frac{x^4+x^2+1}{x^4+1}\)
\(\frac{x^2+y^2-4+2xy}{x^2-y^2+4+4x}\)
\(=\frac{\left(x+y\right)^2-2^2}{\left(x+2\right)^2-y^2}\)
\(=\frac{\left(x+y-2\right)\left(x+y+2\right)}{\left(x+2-y\right)\left(x+2+y\right)}\)
\(=\frac{x+y-2}{x+2-y}\)
\(\frac{4x^2+12x+9}{2x^2-x-6}\)
\(=\frac{\left(2x+3\right)^2}{2x^2-4x+3x-6}\)
\(=\frac{\left(2x+3\right)^2}{2x\left(x-2\right)+3\left(x-2\right)}\)
\(=\frac{\left(2x+3\right)^2}{\left(2x+3\right)\left(x-2\right)}\)
\(=\frac{2x+3}{x-2}\)
\(\frac{25-10x+x^2}{xy-5y}\)
\(=\frac{\left(5-x\right)^2}{-y\left(5-x\right)}\)
\(=-\frac{5-x}{y}\)
\(\frac{\left|x\right|-3}{x^2-9}\)
\(=\frac{x-3}{\left(x+3\right)\left(x-3\right)}\)
\(=\frac{1}{x+3}\)
\(\frac{3\left|x-4\right|}{3x^2-3x-36}\)
\(=\frac{3\left(x-4\right)}{3\left(x^2-x-12\right)}\)
\(=\frac{x-4}{x^2-4x+3x-12}\)
\(=\frac{x-4}{x\left(x-4\right)+3\left(x-4\right)}\)
\(=\frac{x-4}{\left(x-4\right)\left(x+3\right)}\)
\(=\frac{1}{x+3}\)
Ta có: \(\frac{x^2y+2xy^2+y^3}{2x^2+xy-y^2}\)
\(=\frac{x^2y+xy^2+xy^2+y^3}{2x^2+2xy-xy-y^2}\)
\(=\frac{xy\left(x+y\right)+y^2\left(x+y\right)}{2x\left(x+y\right)-y\left(x+y\right)}\)
\(=\frac{\left(x+y\right)\left(xy+y^2\right)}{\left(2x-y\right)\left(x+y\right)}=\frac{xy+y^2}{2x-y}\left(đpcm\right)\)
Ta có: \(\frac{x^2+3xy+2y^2}{x^3+2x^2y-xy^2-2y^3}\)
\(=\frac{x^2+xy+2xy+2y^2}{x^2\left(x+2y\right)-y^2\left(x+2y\right)}\)
\(=\frac{x\left(x+y\right)+2y\left(x+y\right)}{\left(x^2-y^2\right)\left(x+2y\right)}\)
\(=\frac{\left(x+2y\right)\left(x+y\right)}{\left(x+y\right)\left(x-y\right)\left(x+2y\right)}=\frac{1}{x-y}\left(đpcm\right)\)
a) (2x^2 +2xy - xy -y^2 ) / (2x^2 - 2xy - xy +y^2)
= 2x(x+y) - y(x+y) / 2x(x-y) - y(x-y)
= (2x-y)(x+y) / (2x-y)(x-y)
= x+y/x-y
Rút gọn cái sau:
\(\frac{32x+4x^2+2x^3}{x^3+64}\)
\(=\frac{2x\left(x^2+2x+16\right)}{\left(x+4\right)\left(x^2-4x+16\right)}\)
Đề có vẻ sai sai ?
a) \(\frac{x^2-xy-x+y}{x^2+xy-x-y}\)=\(\frac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)=\(\frac{\left(x-1\right)\left(x-y\right)}{\left(x-1\right)\left(x+y\right)}\)=\(\frac{x-y}{x+y}\)
b) \(\frac{x^2-xy}{5y^2-5xy}\)=\(\frac{x\left(x-y\right)}{-5y\left(x-y\right)}\)=\(\frac{-x}{5y}\)
c) \(\frac{3x^2-12x+12}{x^4-8x}\)=\(\frac{3\left(x^2-4x+4\right)}{x\left(x^3-2^3\right)}\)=\(\frac{3\left(x-2\right)^2}{x\left(x-2\right)\left(x^2+2x+4\right)}\)=\(\frac{3\left(x-2\right)}{x\left(x^2+2x+4\right)}\)