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\(x^2-xy-8x+8y\)
\(=x\left(x-y\right)-8\left(x-y\right)=\left(x-8\right)\left(x-y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài làm
8x2 - 12xy - 8y2 = 0
=> ( 8x2 - 8y2 ) - 12xy = 0
=> 8( x2 - y2 ) - 12xy = 0
=> 8( x - y )( x + y ) - 12xy = 0
=> 4[ 2( x - y )( x + y ) - 3xy ] = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^2-3x=x\left(x-3\right)\)
b) \(10x\left(x-y\right)-8y\left(x-y\right)=2\left(x-y\right)\left(5x-4y\right)\)
c) \(x^2-9=\left(x-3\right)\left(x+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(14x^2-14xy-8x+8y=14x\left(x-y\right)-8\left(x-y\right)=\left(x-y\right)\left(14x-8\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(4x^2-y^2+8x-16\)
\(=\left(2x\right)^2-\left(y-4\right)^2=\left(2x-y+4\right)\left(2x+y-4\right)\)
4x2 - y2 + 8y - 16
= 4x2 - (y2 - 8y + 16)
= (2x)2 - (y - 4)2
= [2x - (y - 4)][2x + (y - 4)]
= (2x - y +4)(2x + y - 4)
![](https://rs.olm.vn/images/avt/0.png?1311)
D = x2 - 4x - y2 - 8y - 12
= (x2 - 4x + 4) - (y2 + 8y + 16)
= (x - 2)2 - (y + 4)2
= (x + y + 2)(x - y - 6)
\(D=x^2-4x-y^2-8y-12\)
\(=x^2-4x-y^2-8y+4-16\)
\(=\left(x^2-4x+4\right)-\left(y^2+8y+16\right)\)
\(=\left(x-2\right)^2-\left(y+4\right)^2\)
\(=\left(x-2-y-4\right)\left(x-2+y+4\right)\)
\(=\left(x-y-6\right)\left(x+y+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a
\(8x^3-\dfrac{1}{125}y^3\\ =\left(2x\right)^3-\left(\dfrac{1}{5}y\right)^3\\ =\left(2x-\dfrac{1}{5}y\right)\left[\left(2x\right)^2+2x.\dfrac{1}{5}y+\left(\dfrac{1}{5}y\right)^2\right]\\ =\left(2x-\dfrac{1}{5}y\right)\left(4x^2+\dfrac{2}{5}xy+\dfrac{1}{25}y^2\right)\)
b
\(-x^3+6x^2y-12xy^2+8y^3\\ =-\left(x^3-6x^2y+12xy^2-8y^3\right)\\ =-\left(x^3-3.2y.x^2+3.\left(2y\right)^2.x-\left(2y\right)^3\right)\\ =-\left(x-2y\right)^3\\ =-\left(x-2y\right)\left(x-2y\right)\left(x-2y\right)\)
a: 8x^3-1/125y^3
=(2x)^3-(1/5y)^3
=(2x-1/5y)(4x^2+2/5xy+1/25y^2)
b: =(2y-x)^3
![](https://rs.olm.vn/images/avt/0.png?1311)
\(10x\left(x-y\right)-8y\left(y-x\right)\)
\(=10x\left(x-y\right)+8y\left(x-y\right)\)
\(=\left(x-y\right)\left(10x+8y\right)\)
\(=2\left(x-y\right)\left(5x+4y\right)\)
\(x^2-y^2+8x-8y\)
\(=\left(x^2-y^2\right)+\left(8x-8y\right)\)
\(=\left(x+y\right).\left(x-y\right)+8.\left(x-y\right)\)
\(=\left(x-y\right).\left(x+y+8\right)\)
x^2-y^2+8x-8y
=(x^2-y^2)+(8x-8y)
=xy(x-y)+8(x-y)
=(x-y).(xy+8)
sai thì cho mik xin lỗi nhé :(