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\(3x^3-75x=0\Leftrightarrow3x\left(x^2-25\right)=0\Leftrightarrow3x\left(x-5\right)\left(x+5\right)=0\Leftrightarrow x=0;x=-5;x=5\)
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\(x^3+3x^2+2x=0\)
\(\Leftrightarrow x\left(x^2+3x+2\right)=0\)
\(\Leftrightarrow x\left(x+1\right)\left(x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=-2\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^4+3x^3-x-3=0\)
\(\Leftrightarrow x^3\left(x+3\right)-\left(x+3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^3-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+3=0\\x^3-1=0\end{matrix}\right.\) \(\Leftrightarrow\left[{}\begin{matrix}x=-3\\x=1\end{matrix}\right.\)
Vậy \(x\in\left\{-3;1\right\}\)
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a, \(4x^2-4x=-1\Leftrightarrow4x^2-4x+1=0\Leftrightarrow\left(2x-1\right)^2=0\Leftrightarrow x=\frac{1}{2}\)
b, \(27x^3+27x^2+9x+1=0\Leftrightarrow27x^3+1+27x^2+9x=0\)
\(\Leftrightarrow\left(3x+1\right)\left(9x^2-3x+1\right)+9x\left(3x+1\right)=0\)
\(\Leftrightarrow\left(3x+1\right)\left(9x^2+2>0\right)=0\Leftrightarrow x=-\frac{1}{3}\)
c, \(9x^2\left(x+1\right)-4\left(x+1\right)=0\Leftrightarrow\left(9x^2-4\right)\left(x+1\right)=0\)
\(\Leftrightarrow\left(3x-2\right)\left(3x+2\right)\left(x+1\right)=0\Leftrightarrow x=-\frac{2}{3};x=\frac{2}{3};x=-1\)
d, \(\left(x+1\right)^3-25\left(x+1\right)=0\Leftrightarrow\left(x+1\right)\left[\left(x+1\right)^2-25\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-4\right)\left(x+6\right)=0\Leftrightarrow x=-1;x=-6;x=4\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\Leftrightarrow\) \(6x^5-12x^4-17x^4+34x^3-7x^3+14x^2+13x^2-26x-3x+\)6 =0
\(6x^5-29x^4+27x^3+27x^2-29x+6=0\)
\(\Leftrightarrow\left(6x^5-18x^4\right)+\left(-11x^4+33x^3\right)+\left(-6x^3+18x^2\right)+\left(9x^2-27x\right)+\left(-2x+6\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(6x^4-11x^3-6x^2+9x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(\left(6x^4-12x^3\right)+\left(x^3-2x^2\right)+\left(-4x^2+8x\right)+\left(x-2\right)\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(6x^3+x^2-4x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(\left(6x^3+6x^2\right)+\left(-5x^2-5x\right)+\left(x+1\right)\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(6x^2-5x+1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(\left(6x^2-3x\right)+\left(-2x+1\right)\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)\left(x+1\right)\left(2x-1\right)\left(3x-1\right)=0\)
\(\Leftrightarrow x=\left(3;2;-1;\frac{1}{2};\frac{1}{3}\right)\)
3x3 - 27x = 0
=> 3x.(x2 -9) = 0
=> 3x = 0 hoặc x2 - 9 = 0
TH1: 3x = 0
=> x = 0
TH2: x2 - 9 = 0
=> x2 = 9
=> x = 3 hoặc x = -3
Vậy, x \(\in\){ 0; 3; -3}
\(3x^3-27x=0\)
\(3x.x^2-27x=0\)
\(3x.x^2=27x\)
\(x^2=27x:3x\)
\(x^2=9\)
\(x=3\).Vậy x = 3
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