giúp mình rút gọn phân thức
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)
\(\frac{x^2-xy-x+y}{x^2+xy-x-y}\)
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6 tháng 10 2021
a) \(\dfrac{6x^2y^2}{8xy^5}=\dfrac{3x}{4y^3}\)
b) \(=\dfrac{2y}{3\left(x+y\right)^2}=\dfrac{2y}{3x^2+6xy+3y^2}\)
c) \(=\dfrac{2x\left(x+1\right)}{x+1}=2x\)
d) \(=\dfrac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}=\dfrac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}=\dfrac{x-y}{x+y}\)
e) \(=\dfrac{36\left(x-2\right)^3}{-16\left(x-2\right)}=-9\left(x-2\right)^2=-9x^2+36x-36\)
12 tháng 9 2016
Ta có
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)= \(\frac{2y}{3\left(x+y\right)^2}\)
\(\frac{x^2+2x+1}{5x^3+5x^2}=\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}=\frac{x+1}{5x^2}\)
1 tháng 10 2016
a)\(\frac{\left(x+1\right)^2}{5x^2\left(x+1\right)}\)=\(\frac{x+1}{5x^2}\)
b)\(\frac{10y}{15\left(x+y\right)^2}\)
12 tháng 9 2016
Ta có
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)= \(\frac{2y}{3\left(x+y\right)^2}\)
\(\frac{7x^2+14x+7}{3x^2+3x}=\frac{7\left(x+1\right)^2}{3x\left(x+1\right)}=\frac{7\left(x+1\right)}{3x}\)
28 tháng 2 2020
Với đk trên ta có:
P = \(\frac{2}{x}-\left(\frac{x^2}{x^2+xy}+\frac{y^2-x^2}{xy}-\frac{y^2}{xy+y^2}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\left(\frac{x}{x+y}-\frac{\left(x-y\right)\left(x+y\right)}{xy}-\frac{y}{x+y}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\left(\frac{x-y}{x+y}-\frac{\left(x-y\right)\left(x+y\right)}{xy}\right).\frac{x+y}{x^2+xy+y^2}\)
\(=\frac{2}{x}-\frac{x-y}{xy}.\left(xy-\left(x+y\right)^2\right).\frac{1}{x^2+xy+y^2}\)
\(=\frac{2}{x}+\frac{x-y}{xy}\)
\(=\frac{x+y}{xy}\)
\(\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)^3}\)
\(=\frac{10xy^2\left(x+y\right)}{15xy\left(x+y\right)\left(x+y\right)^2}\)
\(=\frac{10y}{15\left(x+y\right)^2}\)
\(\frac{x^2-xy-x+y}{x^2+xy-x-y}\)
\(=\frac{\left(x^2-x\right)-\left(xy-y\right)}{\left(x^2-x\right)+\left(xy-y\right)}\)
\(=\frac{x\left(x-1\right)-y\left(x-1\right)}{x\left(x-1\right)+y\left(x-1\right)}\)
\(=\frac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)
\(=\frac{x-y}{x+y}\)
a)\(\frac{2xy}{3\left(x+y\right)^2}\)
b)=\(\frac{\left(x^2-xy\right)-\left(x-y\right)}{\left(x^2+xy\right)-\left(x+y\right)}\)=\(\frac{x\left(x-y\right)-\left(x-y\right)}{x\left(x+y\right)-\left(x+y\right)}\)
=\(\frac{\left(x-y\right)\left(x-1\right)}{\left(x+y\right)\left(x-1\right)}\)=\(\frac{\left(x-y\right)}{\left(x+y\right)}\)