phân tích đa thức thành nhân tử : x^2-y^2+8x-8y
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\(x^2-xy-8x+8y\)
\(=x\left(x-y\right)-8\left(x-y\right)=\left(x-8\right)\left(x-y\right)\)
\(=x^2-\left(y-4\right)^2\)
\(=\left(x-y+4\right)\left(x+y-4\right)\)
\(=x^2-\left(y^2-8y+16\right)=x^2-\left(y-4\right)^2=\left(x-y+4\right)\left(x+y-4\right)\)
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
\(1,\\ 1,=15\left(x+y\right)\\ 2,=4\left(2x-3y\right)\\ 3,=x\left(y-1\right)\\ 4,=2x\left(2x-3\right)\\ 2,\\ 1,=\left(x+y\right)\left(2-5a\right)\\ 2,=\left(x-5\right)\left(a^2-3\right)\\ 3,=\left(a-b\right)\left(4x+6xy\right)=2x\left(2+3y\right)\left(a-b\right)\\ 4,=\left(x-1\right)\left(3x+5\right)\\ 3,\\ A=13\left(87+12+1\right)=13\cdot100=1300\\ B=\left(x-3\right)\left(2x+y\right)=\left(13-3\right)\left(26+4\right)=10\cdot30=300\\ 4,\\ 1,\Rightarrow\left(x-5\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=5\end{matrix}\right.\\ 2,\Rightarrow\left(x-7\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=7\\x=-2\end{matrix}\right.\\ 3,\Rightarrow\left(3x-1\right)\left(x-4\right)=0\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{3}\\x=4\end{matrix}\right.\\ 4,\Rightarrow\left(2x+3\right)\left(2x-1\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{3}{2}\\x=\dfrac{1}{2}\end{matrix}\right.\)
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) \(x^2-9=\left(x-3\right)\left(x+3\right)\)
\(14x^2-14xy-8x+8y=14x\left(x-y\right)-8\left(x-y\right)=\left(x-y\right)\left(14x-8\right)\)
\(4x^2-y^2+8x-16\)
\(=\left(2x\right)^2-\left(y-4\right)^2=\left(2x-y+4\right)\left(2x+y-4\right)\)
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)
D = x2 - 4x - y2 - 8y - 12
= (x2 - 4x + 4) - (y2 + 8y + 16)
= (x - 2)2 - (y + 4)2
= (x + y + 2)(x - y - 6)
\(D=x^2-4x-y^2-8y-12\)
\(=x^2-4x-y^2-8y+4-16\)
\(=\left(x^2-4x+4\right)-\left(y^2+8y+16\right)\)
\(=\left(x-2\right)^2-\left(y+4\right)^2\)
\(=\left(x-2-y-4\right)\left(x-2+y+4\right)\)
\(=\left(x-y-6\right)\left(x+y+2\right)\)
=(x-y-2y)[(x-y)^2+2y(x-y)+4y^2]
=(x-3y)(x^2-2xy+y^2+2xy-2y^2+4y^2)
=(x-3y)(x^2+3y^2)
\(\left(x-y\right)^3-8y^3\)
\(=\left(x-y\right)^3-\left(2y\right)^3\)
\(=\left[\left(x-y\right)-2y\right]\left[\left(x-y\right)^2+2y\left(x-y\right)+\left(2y\right)^2\right]\)
\(=\left(x-y-2y\right)\left(x^2-2xy+y^2+2xy-2y^2+4y^2\right)\)
\(=\left(x-3y\right)\left(x^2+3y^2\right)\)
\(x^2-y^2+8x+6y+7\)
\(=\left(x-y\right)\left(x+y\right)+7\left(x+y\right)+x-y+7\)
\(=\left(x+y\right)\left(x-y+7\right)+\left(x-y+7\right)\)
\(=\left(x+y+1\right)\left(x-y+7\right)\)
\(x^2-y^2+8x-8y\)
\(=\left(x^2-y^2\right)+\left(8x-8y\right)\)
\(=\left(x+y\right).\left(x-y\right)+8.\left(x-y\right)\)
\(=\left(x-y\right).\left(x+y+8\right)\)
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 :(