Rút gọn: \(\frac{2x^2-4xy}{x^2+4xy+4y^2}:\frac{4y^2-x^2}{x^2-4xy+4y^2}:\frac{5x^2y-10xy^2}{x^3+6x^2y+12xy^2+8y^3}\)
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30 tháng 1 2022
\(=\dfrac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\cdot\dfrac{\left(x-2y\right)^2}{-\left(x-2y\right)\left(x+2y\right)}:\dfrac{5x^2y-10xy^2}{x^3+6x^2y+12xy^3+8y^3}\)
\(=\dfrac{-2x\left(x-2y\right)^2}{\left(x+2y\right)^3}\cdot\dfrac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}\)
\(=\dfrac{-2x\cdot\left(x-2y\right)}{5xy}=\dfrac{-2\left(x-2y\right)}{5y}\)
TN
0
6 tháng 9 2018
1/ x^2 +4xy +4y^2 = (x +2y)^2
2/ -x^3 +9x^2 -27x+27= - (x^3 -9x^2+27x-27) = - (x-3)^3
3/ 8x^6 +36x^4y+54^2y^2+27y^3 = (2x^2+3y)^3
4/ x^3 - 6x^2y+12xy^2 -8y^3= (x-2y)^3
IL
1
29 tháng 9 2019
\(\frac{x^2+3xy+2y^2}{5x^2+4xy-y^2}-\frac{x^2-5xy+4y^2}{-2x^2+4xy-2y^2}\)
\(=\frac{x+2y}{5x-y}-\left[-\frac{x-4y}{2\left(x-y\right)}\right]\)
\(=\frac{x+2y}{5x-y}+\frac{x-4y}{2\left(x-y\right)}\)
\(=\frac{\left(x+2y\right).2\left(x-y\right)}{\left(5x-y\right).2\left(x-y\right)}+\frac{\left(x-4y\right).\left(5x-y\right)}{2\left(x-y\right).\left(5x-y\right)}\)
\(=\frac{\left(x+2y\right).2\left(x-y\right)+\left(x-4y\right).\left(5x-y\right)}{2\left(x-y\right).\left(5x-y\right)}\)
\(=\frac{7x^2-19xy}{2\left(x-y\right).\left(5x-y\right)}\)
17 tháng 10 2021
\(1,=\left(x-2\right)\left(5-y\right)\\ 2,=2\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(2x-2y-z\right)\\ 3,=5xy\left(x-2y\right)\\ 4,=3\left(x^2-2xy+y^2-4z^2\right)=3\left[\left(x-y\right)^2-4z^2\right]\\ =3\left(x-y-2z\right)\left(x-y+2z\right)\\ 5,=\left(x+2y\right)^2-16=\left(x+2y-4\right)\left(x+2y+4\right)\\ 6,=-\left(6x^2-3x-4x+2\right)=-\left(2x-1\right)\left(3x-2\right)\\ 7,=\left(2x+y\right)\left(2x+y+x\right)=\left(2x+y\right)\left(3x+y\right)\\ 8,=\left(x-y\right)\left(x+5\right)\\ 9,=\left(x+1\right)^2-y^2=\left(x-y+1\right)\left(x+y+1\right)\\ 10,=\left(x^2-9\right)x=x\left(x-3\right)\left(x+3\right)\\ 11,=\left(x-2\right)\left(y+1\right)\\ 12,=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\\ 13,=3\left(x+y\right)-\left(x+y\right)^2=\left(x+y\right)\left(3-x-y\right)\)
3 tháng 2 2016
a/=X^4*y*(1-5)+(2-5)*X*Y^3
=X^4*y*-4+-3*X*Y^3
tạm thời tớ giúp được thế thôi.để tớ còn làm nốt ba bài hình cô giao đã
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y-x\right)\left(2y+x\right)}{\left(x-2y\right)^2}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)
Điều kiện: \(x\ne2y;x\ne-2y;x\ne0;y\ne0\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}:\frac{\left(2y+x\right)}{\left(x-2y\right)}:\frac{5xy\left(x-2y\right)}{\left(x+2y\right)^3}\)
\(=\frac{2x\left(x-2y\right)}{\left(x+2y\right)^2}\times\frac{x-2y}{x+2y}\times\frac{\left(x+2y\right)^3}{5xy\left(x-2y\right)}=\frac{2\left(x-2y\right)}{5y}\)