Tìm X < 0 biết [x-2] + 3[2 – x] + [4x -8] = 32 . trả lời X =
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|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
\(\left|x-2\right|+3\left|2-x\right|+\left|4x-8\right|=32\)
\(\Leftrightarrow\left|x-2\right|+3\left|x-2\right|+4\left|x-2\right|=32\)
Đặt \(t=\left|x-2\right|\) ta có:
\(t+3t+4t=32\)\(\Leftrightarrow8t=32\Leftrightarrow t=4\)
\(\left|x-2\right|=4\)\(\Leftrightarrow\left[\begin{matrix}x-2=4\\x-2=-4\end{matrix}\right.\)\(\Leftrightarrow x=-2\left(x< 0\right)\)
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
=>\(\left[{}\begin{matrix}x-2=0\\x-6=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=6\end{matrix}\right.\)