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\(=-12x^2\left(y-x\right)+18x^3\left(y-x\right)\)
\(=-6x^2\left(y-x\right)\left(2-3x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-7x+12\)
\(=\left(x^2-4x\right)-\left(3x-12\right)\)
\(=x\left(x-4\right)-3\left(x-4\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
c ) \(x^2-7x+12\)
\(=\left(x^2-3x\right)-\left(4x-12\right)\)
\(=x\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x-4\right)\left(x-3\right)\)
d ) \(x^2+7x+12\)
\(=\left(x^2+3x\right)+\left(4x+12\right)\)
\(=x\left(x+3\right)+4\left(x+3\right)\)
\(=\left(x+4\right)\left(x+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
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: \(x\left(x-6\right)+10\left(x-6\right)=0\)
\(\Leftrightarrow\left(x-6\right)\left(x+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=6\\x=-10\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) ( x 2 – 4x + 1)( x 2 – 2x + 3).
b) ( x 2 + 5x – 1)( x 2 + x – 1).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(12x^2+7x-12\)
\(=12x^2-9x+16x-12\)
\(=3x\left(4x-3\right)+4\left(4x-3\right)\)
\(=\left(4x-3\right)\left(3x+4\right)\)
12x^2+7x-12=(12x^2-9x)+(16x-12)=3x(4x-3)+4(4x-3)=(3x+4)(4x-3)
chúc hok tốt
![](https://rs.olm.vn/images/avt/0.png?1311)
(x^2+3x+2)(x^2+7x+12)
=(x2+x+2x+2)(x2+3x+4x+12)
=[x.(x+1)+2.(x+1)][x.(x+3)+4.(x+3)]
=(x+1)(x+2)(x+3)(x+4)
12x2+7x-12=-7
Đúng 100%
Mình mới lớp 6 nhưng mình nghĩ thế này nha :
x(12x +7) -12