Question

[Nguyễn ngọc thảo vy]

Giải phương trình sau:

a/ \(\left(x-1\right)^2-1+x^2=\left(1-x\right).\left(x+3\right)\)

b/ \(x^3-7x+6=0\)

c/ \(x^4-4x^3+12x-9=0\)

d/ \(x^5-5x^3+4x=0\)

e/ \(x^4-4x^3+3x^2+4x-4=0\)

Giúp mình với!!!!!!!!!!

Answer 1

[Nguyễn Lê Phước Thịnh]

b) Ta có: \(x^3-7x+6=0\)

\(\Leftrightarrow x^3-6x-x+6=0\)

\(\Leftrightarrow x\left(x^2-1\right)-6\left(x-1\right)=0\)

\(\Leftrightarrow x\left(x-1\right)\left(x+1\right)-6\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[x\left(x+1\right)-6\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+x-6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+3x-2x-6\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+3\right)\left(x-2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+3=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-3\\x=2\end{matrix}\right.\)

Vậy: x∈{1;-3;2}

c) Ta có: \(x^4-4x^3+12x-9=0\)

\(\Leftrightarrow x^4-4x^3+3x^2-3x^2+12x-9=0\)

\(\Leftrightarrow x^2\left(x^2-4x+3\right)-3\left(x^2-4x+3\right)=0\)

\(\Leftrightarrow\left(x^2-4x+3\right)\left(x^2-3\right)=0\)

\(\Leftrightarrow\left(x-3\right)\left(x-1\right)\left(x^2-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-3=0\\x-1=0\\x^2-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=1\\x=\pm\sqrt{3}\end{matrix}\right.\)

Vậy: \(x\in\left\{3;1;\pm\sqrt{3}\right\}\)

d) Ta có: \(x^5-5x^3+4x=0\)

\(\Leftrightarrow x^5-x^3-4x^3+4x=0\)

\(\Leftrightarrow x^3\left(x^2-1\right)-4x\left(x^2-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\left(x^3-4x\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x+1\right)\cdot x\left(x^2-4\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x+1=0\\x=0\\x^2-4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\\x=0\\x=\pm2\end{matrix}\right.\)

Vậy: x∈{-2;-1;0;1;2}

e) Ta có: \(x^4-4x^3+3x^2+4x-4=0\)

\(\Leftrightarrow x^4-4x^3+4x^2-x^2+4x-4=0\)

\(\Leftrightarrow x^2\left(x^2-4x+4\right)-\left(x^2-4x+4\right)=0\)

\(\Leftrightarrow\left(x-2\right)^2\left(x-1\right)\left(x+1\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}\left(x-2\right)^2=0\\x-1=0\\x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x=1\\x=-1\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=1\\x=-1\end{matrix}\right.\)

Vậy: x∈{-1;1;2}