1) PT <=> \(\left(x^3+2x^2\right)-\left(x^2+2x\right)+\left(2x+4\right)=0\)
<=> \(x^2\left(x+2\right)-x\left(x+2\right)+2\left(x+2\right)=0\)
<=> \(\left(x+2\right)\left(x^2-x+2\right)=0\)
<=> \(\left[{}\begin{matrix}x+2=0\\x^2-x+2=0\end{matrix}\right.< =>\left[{}\begin{matrix}x=-2\\x=\varnothing\end{matrix}\right.\)
KL: ...
2) PT <=> \(\left(x^3-x^2\right)-\left(4x^2-4x\right)+\left(4x-4\right)=0\)
<=> \(x^2\left(x-1\right)-4x\left(x-1\right)+4\left(x-1\right)=0\)
<=> \(\left(x-1\right)\left(x^2-4x+4\right)=0\)
<=> \(\left(x-1\right)\left(x-2\right)^2=0\)
<=> \(\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\) <=> \(\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
KL: ...
1)
Sửa đề: \(x^3-3x^2+4=0\)
Ta có: \(x^3-3x^2+4=0\)
\(\Leftrightarrow x^3+x^2-4x^2+4=0\)
\(\Leftrightarrow x^2\left(x+1\right)-4\left(x^2-1\right)=0\)
\(\Leftrightarrow x^2\left(x+1\right)-4\left(x+1\right)\left(x-1\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left[x^2-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x+1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x+1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x+1=0\\\left(x-2\right)^2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)
Vậy: Tập nghiệm S={-1;2}
2) Ta có: \(x^2-5x^2+8x^4=0\)
\(\Leftrightarrow8x^4-4x^2=0\)
\(\Leftrightarrow4x^2\left(4x^2-1\right)=0\)
\(\Leftrightarrow4x^2\left(2x-1\right)\left(2x+1\right)=0\)
mà 4≠0
nên \(\left[{}\begin{matrix}x^2=0\\2x-1=0\\2x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\2x=1\\2x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\frac{1}{2}\\x=\frac{-1}{2}\end{matrix}\right.\)
Vậy: Tập nghiệm \(S=\left\{0;\frac{1}{2};\frac{-1}{2}\right\}\)
