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Ta có: \(2x^3+3x^2+2x+3=0\)
\(\Leftrightarrow x^2\left(2x+3\right)+\left(2x+3\right)=0\)
\(\Leftrightarrow2x+3=0\)
hay \(x=-\dfrac{3}{2}\)
a: Ta có: \(2-x=2\left(x-2\right)^3\)
\(\Leftrightarrow2\left(x-2\right)^3+x-2=0\)
\(\Leftrightarrow\left(x-2\right)\left[2\left(x-2\right)^2+1\right]=0\)
\(\Leftrightarrow x-2=0\)
hay x=2
c: Ta có: \(\left(x-1.5\right)^6+2\left(1.5-x\right)^3=0\)
\(\Leftrightarrow\left(x-1.5\right)^6-2\left(x-1.5\right)^3=0\)
\(\Leftrightarrow\left(x-1.5\right)^3\cdot\left[\left(x-1.5\right)^3-2\right]=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=1.5\\x=\sqrt[3]{2}+1.5\end{matrix}\right.\)
\(\left(x^2+2x\right)^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+4x^2-2x^2-4x-3=0\Leftrightarrow x^4+4x^3+2x^2-4x-3=0\Leftrightarrow\left(x-1\right)\left(x+1\right)^2\left(x-3\right)=0\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=3\end{matrix}\right.\)
Ta có: \(\left(x^2+2x\right)^2-2x^2-4x-3=0\)
\(\Leftrightarrow\left(x^2+2x\right)^2-2\left(x^2+2x\right)-3=0\)
\(\Leftrightarrow\left(x^2+2x-3\right)\left(x^2+2x+1\right)=0\)
\(\Leftrightarrow\left(x+1\right)^2\cdot\left(x+3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-3\\x=1\end{matrix}\right.\)
a) \(27x^3+27x^2+9x+1=\left(3x+1\right)^3\)
b) \(-x^3-3x^2-3x-1=-\left(x^3+3x^2+3x+1\right)=-\left(x+1\right)^3\)
c) \(-8+12x-6x^2+x^3=\left(x-2\right)^3\)
\(x^4+3x^3+12x-16\)
\(=x^4+4x^3+4x^2+16x-x^3-4x^2-4x-16\)
\(=x\left(x^3+4x^2+4x+16\right)-\left(x^3+4x^2+4x+16\right)\)
\(=\left(x-1\right)\left(x^3+4x^2+4x+16\right)\)
\(=\left(x-1\right)\left[x^2\left(x+4\right)+4\left(x+4\right)\right]\)
\(=\left(x-1\right)\left(x+4\right)\left(x^2+4\right)\)
a: \(=3\left(x^2-y^2-x+y\right)\)
\(=3\left[\left(x-y\right)\left(x+y\right)-\left(x-y\right)\right]\)
=3(x-y)(x+y-1)
b: =(x-4)(x+1)
c: =x(x-1)
x^2 - 2x - 15
= x^2 - 5x + 3x - 15
= ( x^2 + 3x ) - (5x +15 )
= x ( x +3 ) - 5 ( x + 3 )
(x + 3 ) ( x - 5 )
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