23x-4+8x=5.62
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![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
d) \(4x^4-x^2=x^2\left(4x^2-1\right)=x^2\left(2x-1\right)\left(2x+1\right)\)
e) Ta có: \(6x^2-7x-5\)
\(=6x^2-10x+3x-5\)
\(=2x\left(3x-5\right)+\left(3x-5\right)\)
\(=\left(3x-5\right)\left(2x+1\right)\)
f: Ta có: \(-4x^2+23x-15\)
\(=-4x^2+20x+3x-15\)
\(=-4x\left(x-5\right)+3\left(x-5\right)\)
\(=\left(x-5\right)\left(-4x+3\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow x^4-4x^3+4x^2-4x^3+16x^2-16x+3x^2-12x+12\le0\)
\(\Leftrightarrow x^2\left(x^2-4x+4\right)-4x\left(x^2-4x+4\right)+3\left(x^2-4x+4\right)\le0\)
\(\Leftrightarrow\left(x^2-4x+3\right)\left(x-2\right)^2\le0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=2\\x^2-4x+3\le0\end{matrix}\right.\) \(\Rightarrow1\le x\le3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x=-2+\sqrt{5}>0\Rightarrow x+2=\sqrt{5}\)
\(\Rightarrow\left(x+2\right)^2=5\Rightarrow x^2+4x=1\)
Ta có:
\(3x^5+12x^4-8x^3-23x^2-7x+1\)
\(=3x^3\left(x^2+4x\right)-8x^3-23x^2-7x+1\)
\(=-5x^3-23x^2-7x+1=-5x\left(x^2+4x\right)-3x^2-7x+1\)
\(=-3x^2-12x+1=-3\left(x^2+4x\right)+1=-3+1=-2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
8x - 10x - 23x = 25
(8-10-23). x = 25
(-25) . x = 25
x = 25 : (-25)
x = -1
Vậy x =-1
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(=16x^2-56x\)
b) \(=161x^3+184x^2-207x\)
c) \(=7x^3y^2+28x^2y^3-7x^2y^2+7xy^3\)
d) \(=-5x^3+10x^2+5x\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có: \(f\left(x\right)=x^4+8x^3+23x^2+28x+12\)
=> \(f\left(x\right)=x^4+3x^3+5x^3+15x^2+8x^2+24x+4x+12\)
=> \(f\left(x\right)=x^3\left(x+3\right)+5x^2\left(x+3\right)+8x\left(x+3\right)+4\left(x+3\right)\)
=> \(f\left(x\right)=\left(x+3\right)\left(x^3+5x^2+8x+4\right)\)
=> \(f\left(x\right)=\left(x+3\right)\left(x^3+2x^2+3x^2+6x+2x+4\right)\)
=> \(f\left(x\right)=\left(x+3\right)\left[x^2\left(x+2\right)+3x\left(x+2\right)+2\left(x+2\right)\right]\)
=> \(f\left(x\right)=\left(x+3\right)\left(x+2\right)\left(x^2+3x+2\right)\)
=> \(f\left(x\right)=\left(x+2\right)\left(x+3\right)\left(x^2+2x+x+2\right)\)
=> \(f\left(x\right)=\left(x+3\right)\left(x+2\right)\left[x\left(x+2\right)+\left(x+2\right)\right]\)
=> \(f\left(x\right)=\left(x+3\right)\left(x+2\right)^2\left(x+1\right)\)