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1)
\(y^2-4y+4-x^2\\ =\left(y-2\right)^2-x^2\\ =\left(y-2-x\right)\left(y-2+x\right)\)
2)
\(8x^3-12x^2+6x-2\\ =2\left(4x^3-6x^2+3x-1\right)\\ =2\left(4x^3-4x^2-2x^2+2x+x-1\right)\\ =2\left(4x^2\left(x-1\right)-2x\left(x-1\right)+\left(x-1\right)\right)\\ =2\left(x-1\right)\left(4x^2-2x+1\right)\)
1) \(y^2-4y+4-x^2\)
\(=\left(y^2-4y+4\right)-x^2\)
\(=\left(y-2\right)^2-x^2\)
\(=\left(y-2-x\right)\left(y-2+x\right)\)
2) \(8x^3-12x^2+6x-1\)
\(=\left(2x\right)^3-3\cdot\left(2x\right)^2\cdot1+3\cdot2x\cdot1^2-1^3\)
\(=\left(2x-1\right)^3\)
\(=\left(2x-1\right)\left(2x-1\right)\left(2x-1\right)\)
1.
\(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\\ =\left(12x^2+6x\right)\left(y+z+y-z\right)\\ =2y\left(12x^2+6x\right)\\ =2y.6x\left(2x+1\right)\\ =12xy\left(2x+1\right)\)
2.
\(x\left(x-6\right)+10\left(x-6\right)=0\\ \Leftrightarrow\left(x-6\right)\left(x+10\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
Vậy \(x\in\left\{6;-10\right\}\) là nghiệm của pt
Bài 1:
Ta có: \(\left(12x^2+6x\right)\left(y+z\right)+\left(12x^2+6x\right)\left(y-z\right)\)
\(=\left(12x^2+6x\right)\left(y+z+y-z\right)\)
\(=6x\left(2x+1\right)\cdot2y\)
\(=12xy\left(2x+1\right)\)
Bài 2:
Ta có: \(x\left(x-6\right)+10\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
a, 2xy^2 ( x^3 -3xy - 4 )
b, x^2 - 4x - 4x +16
= x(x-4) - 4(x-4)
= (x-4) (x-4)
a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
1, \(xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(x+y\right)\)
2, \(5x\left(3y+4x-6\right)\)
3, \(3x\left(2-y\right)\)
4, \(x\left(x^2+2x+1\right)=x\left(x+1\right)^2\)
5, \(x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\)
6, \(2xy\left(x+2y-5x^2y\right)\)
7, \(x^2\left(x^2+2x+1\right)=x^2\left(x+1\right)^2\)
11, \(\left(x+y\right)\left(x-1\right)\)
\(1,xy^3-x^3y=xy\left(y^2-x^2\right)=xy\left(y-x\right)\left(y+x\right)\\ 2,15xy+20x^2-30x=5x\left(3y+4x-6\right)\\ 3,6x-3xy=3x\left(2-y\right)\\ 4,x^3+2x^2+x=x\left(x^2+2x+1\right)=x\left(x+1\right)^2\\ 5,4x^3-12x^2+9x=x\left(4x^2-12x+9\right)=x\left(2x-3\right)^2\\ 6,2x^2y+4xy^2-10x^3y^2=2xy\left(x+2y-5x^2y\right)\\ 11,x\left(x-1\right)-y\left(1-x\right)=x\left(x-1\right)+y\left(x-1\right)=\left(x-1\right)\left(x+y\right)\)
\(8x^3-\frac{1}{8}\)
\(=\left(2x\right)^3-\left(\frac{1}{2}\right)^3\)
\(=\left(2x-\frac{1}{2}\right)\left(4x^2+x+\frac{1}{4}\right)\)
tích hộ
\(2x^3-12x^2+18x=2x\left(x^2-6x+9\right)=2x\left(x-3\right)^2\)
\(8x^3+12x^2+6x+1\)
\(=\left(2x\right)^3-3\left(2x\right)^2.1+2.3x.1^2-1^3\)
\(=\left(2x-1\right)^3\)
(2x+1)3 nhé mn bn này lm sai ạ