Phân tích đa thức thành nhân tử
a^3x - ab + b - x
giải nhanh giúp với
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
1a) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
b) \(=-\left(x^3-3x^2+3x-1\right)=-\left(x-1\right)^3\)
\(a,=-\left(x-1\right)^3\left[=\left(1-x\right)^3\right]\\ b,=\left(1-x\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(x^3+3x^2+3x=0\Rightarrow x\left(x^2+3x+3\right)=0\Rightarrow x\left[\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\right]=0\Rightarrow x=0\)
(do \(\left(x+\dfrac{3}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4}>0\))
b) \(x^3+6x^2+12x=0\Rightarrow x\left(x^2+6x+12\right)=0\Rightarrow x\left[\left(x+3\right)^2+4\right]=0\Rightarrow x=0\)
(do (x+3)2+4≥4>0)
a: Ta có: \(x^3+3x^2+3x=0\)
\(\Leftrightarrow x\left(x^2+3x+3\right)=0\)
hay x=0
b: Ta có: \(x^3+6x^2+12x=0\)
\(\Leftrightarrow x\left(x^2+6x+12\right)=0\)
hay x=0
![](https://rs.olm.vn/images/avt/0.png?1311)
1/(x+2)2 -(3x-1)2=(x+2+3x-1)(x+2-3x+1)=4x(-2x+3)=-8x2+12x
2/(x4+x2)(-2x3-2x)=x2(x2+1)-2x(x2+1)=(x2+1)(x2-2x)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(ab\left(a-b\right)-ac\left(a+c\right)+bc\left(2a-b+c\right)\)
\(=ab\left(a-b\right)-ac\left(a+c\right)+bc\left[\left(a-b\right)+\left(a+c\right)\right]\)
\(=ab\left(a-b\right)-ac\left(a+c\right)+bc\left(a-b\right)+bc\left(a+c\right)\)
\(=\left(a-b\right)\left(ab+bc\right)+\left(a+c\right)\left(bc-ac\right)\)
\(=b\left(a-b\right)\left(a+c\right)-c\left(a+c\right)\left(a-b\right)\)
\(=\left(b-c\right)\left(a-b\right)\left(a+c\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(ab+a+b+1\)
\(=a\left(b+1\right)+\left(b+1\right)\)
\(=\left(b+1\right)\left(a+1\right)\)
c: \(4x^2-12xy+3x-9y\)
\(=4x\left(x-3y\right)+3\left(x-3y\right)\)
\(=\left(x-3y\right)\left(4x+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có :
\(x^4-3x^2+1\)
\(=\left(x^4-2x^2+1\right)-x^2\)
\(=\left(x^2-1\right)^2-x^2\)
\(=\left(x^2-1-x\right)\left(x^2-1+x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^3-3x^2+3x-1=x^3+x^3-3x^2+3x-1\)
=\(x^3+\left(x^3-3x^2+3x-1\right)\)=\(x^3+\left(x-1\right)^3\)
=\(\left(x+x-1\right)\left(x^2-x\left(x-1\right)+\left(x-1\right)^2\right)\)
=\(\left(2x-1\right)\left(x^2-x^2+x+x^2-2x+1\right)\)
=\(\left(2x-1\right)\left(x^2-x+1\right)\)
= (a3x - x) - ( ab - b ) = x (a3 - 1 ) - b(a - 1 ) = x (a - 1) (a2 + a + 1) - b(a - 1 ) = (a - 1) ( a2x + ax + x - b)