3x-12x2 đc phân tích thành ( tích tối đa )
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\(2x^3-12x^2+18x=2x\left(x^2-6x+9\right)=2x\left(x-3\right)^2\)
\(=-12x^2\left(y-x\right)+18x^3\left(y-x\right)\)
\(=-6x^2\left(y-x\right)\left(2-3x\right)\)
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) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
Câu 1:
a) 2x(3x+2) - 3x(2x+3) = 6x^2+4x - 6x^2-9x = -5x
b) \(\left(x+2\right)^3+\left(x-3\right)^2-x^2\left(x+5\right)\)
\(=x^3+6x^2+12x+8+x^2-6x+9-x^3-5x^2\)
\(=2x^2+6x+17\)
c) \(\left(3x^3-4x^2+6x\right)\div\left(3x\right)=x^2-\dfrac{4}{3}x+2\)
`3x-12x^{2}`
`=3x.1-3x.4x`
`=3x.(1-4x)`
Lời giải:
$3x-12x^2=3x(1-4x)$