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a) x3+4x2+x-6=0
<=> x3+x2-2x+3x2+3x-6=0
<=>x(x2+x-2)+3(x2+x-2)=0
<=>(x+3)(x2+x-2)=0
<=>(x+3)(x2+2x-x-2)=0
<=>(x+3)[x(x+2)-(x+2)]=0
<=>(x+3)(x-1)(x+2)=0
=> x+3=0 hay
x-1=0 hay
x+2=0
<=> x=-3 hay x=1 hay x=-2
b)x3-3x2+4=0
\(\Leftrightarrow x^3-4x^2+4x+x^2-4x+4=0\)
\(\Leftrightarrow x\left(x^2-4x+4\right)+\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)^2=0\)
\(\Rightarrow\left\{\begin{matrix}x+1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left\{\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
a: \(\left(3x-1\right)^2-\left(x+3\right)^3=\left(2-x\right)\left(x^2+2x+4\right)\)
\(\Leftrightarrow9x^2-6x+1-x^3-9x^2-27x-27=8-x^3\)
\(\Leftrightarrow-x^3-33x-26-8+x^3=0\)
=>-33x=34
hay x=-34/33
b: \(\left(x+1\right)\left(x-1\right)\left(x^2+1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow\left(x^2+1\right)\left(x^2-1\right)-\left(x^2-1\right)^2=2\)
\(\Leftrightarrow x^4-1-x^4+2x^2-1=2\)
\(\Leftrightarrow2x^2=4\)
hay \(x\in\left\{\sqrt{2};-\sqrt{2}\right\}\)
c: \(x^2-2\sqrt{3}x+3=0\)
\(\Leftrightarrow\left(x-\sqrt{3}\right)^2=0\)
hay \(x=\sqrt{3}\)
d: \(\left(x-\sqrt{2}\right)\left(x+\sqrt{2}\right)-\left(x-\sqrt{2}\right)^2=0\)
\(\Leftrightarrow\left(x-\sqrt{2}\right)\left(x+\sqrt{2}-x+\sqrt{2}\right)=0\)
\(\Leftrightarrow x-\sqrt{2}=0\)
hay \(x=\sqrt{2}\)
\(x^3-6x^2+5x+12>0\\ < =>\left(x^3-5x-x+5x\right)+12>0\\ < =>\left[\left(x^3-x\right)-\left(5x-5x\right)\right]+12>0\\ < =>x^2+12>0\\ < =>x^2>-12\\ =>x\in R\\ BPTcóvôsốnghiem\)
a) (x+2)2 - (x-2).(x+2) = 0
=> x2+4x+4-x2+4 = 0
=> 4x+8 = 0
=> 4x = 0-8
=> 4x = -8
=> x = -8:4
=> x = -2
c) x=0
tíck cho mình nha