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\(a,\frac{4x^3}{10x^2y}=\frac{2x}{5y}\)
\(b,\frac{10xy^5\left(2x-3y\right)}{12xy\left(2x-3y\right)}=\frac{5y^4}{6}\)
Hok Tốt~~
\(\frac{4x^3}{10x^2y}=\frac{2x}{5y}\)
\(\frac{10xy^5\left(2x-3y\right)}{12xy\left(2x-3y\right)}=\frac{5y^4}{4}\)
Tham khảo nhé~
![](https://rs.olm.vn/images/avt/0.png?1311)
a)\(\frac{x^2+y^2-1+2xy}{x^2-y^2+1+2x}\)
\(\Leftrightarrow\frac{\left(x+y\right)^2-1}{\left(x+1\right)^2-y^2}\)
\(\Leftrightarrow\frac{\left(x+y+1\right)\left(x+y-1\right)}{\left(x+1-y\right)\left(x+1+y\right)}\)
\(\Leftrightarrow\frac{x+y-1}{x-y+1}\)
b)\(\frac{3x^3-6x^2y+xy^2-2y^3}{9x^5-18x^4y-xy^4+2y^5}\)
\(\Leftrightarrow\frac{3x^2\left(x-2y\right)+y^2\left(x-2y\right)}{9x^4\left(x-2y\right)-y^4\left(x-2y\right)}\)
\(\Leftrightarrow\frac{\left(3x^2+y^2\right)\left(x-2y\right)}{\left(9x^4-y^4\right)\left(x-2y\right)}\)
\(\Leftrightarrow\frac{3x^2+y^2}{\left(3x^2-y^2\right)\left(3x^2+y^2\right)}\)
\(\Leftrightarrow\frac{1}{3x^2-y^2}\)
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\(\frac{14xy^5\left(2x-3y\right)}{21x^2y\left(2x-3y\right)^2}=\frac{2y^4}{3x\left(2x-3y\right)}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=\frac{2.7.xy.y^5\left(2x-3y\right)}{3.7.xy.x\left(2x-3y\right)^2}=\frac{2y^5}{3x\left(2x-3y\right)}\)
\(\frac{14xy^5\left(2x-3y\right)}{21x^2y\left(2x-3y\right)^2}\)
\(=\frac{2y^4}{3x\left(2x-3y\right)}\)