x âm = X ÂM

|x-2|+3|2-x|+|4x-8|=32

|x-2|+3|x-2|+4|x-2|=32

|x-2|(1+3+4)=32

|x-2|.8=32

|x-2|=4

=> x-2=4 hoac x-2=-4

=> x=6.       ,    x=-2

mà x âm => x=-2

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????????

\(\left|x-2\right|+3\left|2-x\right|+\left|4x-8\right|=32\)

\(\Leftrightarrow\left|x-2\right|+3\left|x-2\right|+4\left|x-2\right|=32\)

Đặt \(t=\left|x-2\right|\) ta có:

\(t+3t+4t=32\)\(\Leftrightarrow8t=32\Leftrightarrow t=4\)

\(\left|x-2\right|=4\)\(\Leftrightarrow\left[\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\)\(\Leftrightarrow x=-2\left(x< 0\right)\)

lưu ý |2-x|=|2-x| rút ra đjăt ẩn..

a: \(x^3-4x^2-x+4=0\)

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

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

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

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

b: Sửa đề: \(x^3+3x^2+3x+1=0\)

=>\(x^3+3\cdot x^2\cdot1+3\cdot x\cdot1^2+1^3=0\)

=>\(\left(x+1\right)^3=0\)

=>x+1=0

=>x=-1

c: \(x^3+3x^2-4x-12=0\)

=>\(\left(x^3+3x^2\right)-\left(4x+12\right)=0\)

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

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

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

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

d: \(\left(x-2\right)^2-4x+8=0\)

=>\(\left(x-2\right)^2-\left(4x-8\right)=0\)

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

=>\(\left(x-2\right)\left(x-2-4\right)=0\)

=>(x-2)(x-6)=0

=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)