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a: \(\Leftrightarrow x\left(x-5\right)\left(x+5\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=5\\x=-5\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)-x\left(x^2-3\right)=0\)
\(\Leftrightarrow x^3-27-x^3+3x=0\)
\(\Leftrightarrow x=9\)
b: Ta có: \(8x^4+x=0\)
\(\Leftrightarrow x\left(8x^3+1\right)=0\)
\(\Leftrightarrow x\left(2x+1\right)\left(4x^2-2x+1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-\dfrac{1}{2}\end{matrix}\right.\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) x = 1; x = - 1 3 b) x = 2.
c) x = 3; x = -2. d) x = -3; x = 0; x = 2.
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Bài 2:
a: Ta có: \(A=\left(x+1\right)^3+\left(x-1\right)^3\)
\(=x^3+3x^2+3x+1+x^3-3x^2+3x-1\)
\(=2x^3+6x\)
b: Ta có: \(B=\left(x-3\right)^3-\left(x+3\right)\left(x^2-3x+9\right)+\left(3x-1\right)\left(3x+1\right)\)
\(=x^3-9x^2+27x-27-x^3-27+9x^2-1\)
\(=27x-55\)
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\(a,\Leftrightarrow6x^2-6x^2-11x+10=-12\\ \Leftrightarrow-11x=-22\\ \Leftrightarrow x=2\\ b,\Leftrightarrow x^3+27-x^3-2x=12-5x\\ \Leftrightarrow3x=-15\\ \Leftrightarrow x=-5\\ c,\Leftrightarrow x^2-6x-16=0\\ \Leftrightarrow\left(x-8\right)\left(x+2\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=8\\x=-2\end{matrix}\right.\)
a: ta có: \(6x^2-\left(2x+5\right)\left(3x-2\right)=-12\)
\(\Leftrightarrow6x^2-6x^2+4x-15x+10=-12\)
\(\Leftrightarrow-11x=-22\)
hay x=2
b: Ta có: \(\left(x+3\right)\left(x^2-3x+9\right)-x\left(x^2+2\right)=12-5x\)
\(\Leftrightarrow x^3+27-x^3-2x+5x=12\)
\(\Leftrightarrow x=-5\)
\(\left(x-3\right)\left(x^2+3x+9\right)-\left(3x-17\right)=x^3-12\)
\(x^3+3x^2+9x-3x^2-9x-37-3x+17=x^3-12\)
\(x^3+3x^2+9x-3x^2-9x-3x-x^3=37-17+12\)
\(-3x=32\)
\(x=\frac{32}{-3}=-\frac{32}{3}\)
Vậy x = \(-\frac{32}{3}\)
(x-3)(x\(^2\)+3x+9)-(3x-17)=x\(^3\)-12
(x\(^3\)-3\(^3\))-(3x-17)=x\(^3\)-12
3\(^3\)-3\(^3\)-3x+17=x\(^3\)-12
x\(^3\)-27-3x+17=x\(^3\)-12
x\(^3\)-27-3x+17-x\(^3\)=-12
-27-3x+17=-12
-27-3x=-29
3x=2
x=\(\frac{2}{3}\)