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\(b,\left(x+2\right)\left(x^2-2x+4\right)-x\left(x^2+2\right)=15\)
\(\Leftrightarrow x^3+8-x^3-2x=15\)
\(\Leftrightarrow-2x=15-8=7\)
\(\Leftrightarrow x=\frac{-7}{2}\)
Vậy \(x=\frac{-7}{2}\)
a: \(\Leftrightarrow x^3-27-x\left(x^2-4\right)=1\)
\(\Leftrightarrow x^3-27-x^3+4x=1\)
=>4x-27=1
hay x=7
b: \(\Leftrightarrow x^3-9x^2+27x-27-x^3+27+6\left(x+1\right)^2+3x^2=15\)
\(\Leftrightarrow-9x^2+27x+6x^2+12x+6+3x^2=15\)
=>39x+6=15
hay x=3/13
c: \(\Leftrightarrow x^3-3x^2+3x-1-x^3-27+3x^2-12=2\)
\(\Leftrightarrow3x-40=2\)
hay x=14
1. \(x^2\left(x+1\right)+x+1=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2+1\right)=0\)
\(\Leftrightarrow x+1=0\Rightarrow x=-1\)
2. \(\left(x-2\right)\left(6x+2\right)+\left(x-2\right)^2=0\)
\(\Leftrightarrow\left(x-2\right)\left(6x+2+x-2\right)=0\)
\(\Leftrightarrow\left(x-2\right).7x=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\7x=0\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}x=2\\x=0\end{matrix}\right.\)
3.
\(x^2-5x+6=0\)
\(\Leftrightarrow x^2-2x-3x+6=0\)
\(\Leftrightarrow x\left(x-2\right)-3\left(x-2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-2\right)=0\Rightarrow\left[{}\begin{matrix}x=2\\x=3\end{matrix}\right.\)
4.
\(x^2-x-6=0\)
\(\Leftrightarrow x^2+2x-3x-6=0\)
\(\Leftrightarrow x\left(x+2\right)-3\left(x+2\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x+2\right)=0\Rightarrow\left[{}\begin{matrix}x=3\\x=-2\end{matrix}\right.\)
b) Ta có: \(\left(x-2\right)\left(x^2-2x+4\right)\left(x+2\right)\left(x^2+2x+4\right)-x^6+2x=1\)
\(\Leftrightarrow\left(x^3-8\right)\left(x^3+8\right)-x^6+2x-1=0\)
\(\Leftrightarrow x^6-64-x^6+2x-1=0\)
\(\Leftrightarrow2x-65=0\)
\(\Leftrightarrow2x=65\)
hay \(x=\frac{65}{2}\)
Vậy: \(x=\frac{65}{2}\)
c) Ta có: \(\left(x-3\right)\left(x^2+3x+9\right)+x\left(x+2\right)\left(2-x\right)=1\)
\(\Leftrightarrow x^3-27-x\left(x+2\right)\left(x-2\right)-1=0\)
\(\Leftrightarrow x^3-27-x\left(x^2-4\right)-1=0\)
\(\Leftrightarrow x^3-27-x^3+4x-1=0\)
\(\Leftrightarrow4x-28=0\)
\(\Leftrightarrow4x=28\)
hay x=7
Vậy: x=7
\(\left(x-2\right)^3-x^2\left(x-6\right)=4\)
\(x^3-6x^2+12x-8-x^3+6x^2=4\)
\(12x-8=4\)
\(12x=4+8\)
\(12x=12\)
\(\Rightarrow x=1\)
Vậy \(x=1\)
\(\left(x+1\right)^3-x\left(x-2\right)^2+x-1=0\)
\(x^3+3x^2+3x+1-x^3+4x^2-4x+x-1=0\)
\(7x^2=0\)
\(\Rightarrow x=0\)
Vậy \(x=0\)
Tham khảo nhé~