Question

Bài 1; Tìm x, biết:

a) \(\text{2. |4 - x| = |-8|}\)

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

c) \(x\left(x-1\right)=0\)

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

Answer 1

a) \(2.\left|4-x\right|=\left|-8\right|\Leftrightarrow2.\left|4-x\right|=8\)

th1: \(4-x\ge0\Leftrightarrow x\le4\)

\(\Rightarrow2.\left|4-x\right|=8\Leftrightarrow2.\left(4-x\right)=8\Leftrightarrow8-2x=8\Leftrightarrow-2x=0\Leftrightarrow x=0\left(tmđk\right)\)

th2: \(4-x< 0\Leftrightarrow x>4\)

\(\Rightarrow2.\left|4-x\right|=8\Leftrightarrow2.\left(x-4\right)=8\Leftrightarrow2x-8=8\Leftrightarrow2x=16\Leftrightarrow x=8\left(tmđk\right)\)

vậy \(x=0;x=8\)

b) đề sai nha bn ; xữa đề : \(\left(x-1\right)^2=0\Leftrightarrow x-1=0\Leftrightarrow x=1\) vậy \(x=1\)

c) \(x\left(x-1\right)=0\Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\left\{{}\begin{matrix}x=0\\x=1\end{matrix}\right.\) vậy \(x=0;x=1\)

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

vậy \(x=-1;x=2\)

Answer 2

a) \(2.\left|4-x\right|=\left|-8\right|\)

\(\Rightarrow2.\left|4-x\right|=8\)

\(\Rightarrow\left[{}\begin{matrix}2\left(4-x\right)=8\\2\left(x-4\right)=8\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}4-x=4\\x-4=4\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=0\\x=8\end{matrix}\right.\)

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

\(\Leftrightarrow x-1=0\)

\(\Leftrightarrow x=1\)

Vậy \(x=1\)

c) \(x\left(x-1\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x-1=0\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Vậy \(\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

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

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

Vậy \(\left[{}\begin{matrix}x=-1\\x=2\end{matrix}\right.\)