Câu hỏi của Trần Thiên Kim

Question

Giải PT:

1. |x2-1|+|x-1|=0

2. |x2-1|+|2x-2|+(x-1)2=0

3. |x-2|+|3x-3|=2x-1

ngonhuminh · Accepted Answer

(1)

$\left\{{}\begin{matrix}\left|x^2-1\right|\ge0\\left|x-1\right|\ge0\end{matrix}\right.$ => để (1) có nghiệm $\left\{{}\begin{matrix}x^2-1=0\x-1=0\end{matrix}\right.$=> x=1

(2)

tương tự $\left\{{}\begin{matrix}x^2-1=0\2x-2=0\x-1=0\end{matrix}\right.$ => x=1Câu trả lời: (1)

$\left\{{}\begin{matrix}\left|x^2-1\right|\ge0\\left|x-1\right|\ge0\end{matrix}\right.$ => để (1) có nghiệm $\left\{{}\begin{matrix}x^2-1=0\x-1=0\end{matrix}\right.$=> x=1

(2)

tương tự $\left\{{}\begin{matrix}x^2-1=0\2x-2=0\x-1=0\end{matrix}\right.$ => x=1

Lightning Farron · Answer

1.$\left|x^2-1\right|+\left|x-1\right|=0$

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

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

$\Leftrightarrow\left[{}\begin{matrix}\left|x-1\right|=0\\left|x+1\right|=1\end{matrix}\right.$$\Leftrightarrow x=1$ (thỏa mãn)

2.$\left|x^2-1\right|+\left|2x-2\right|+\left(x-1\right)^2=0$

$\Leftrightarrow\left[{}\begin{matrix}\left|x-1\right|=0\\left|x+1\right|+2+\left|x-1\right|=0\end{matrix}\right.$

$\Leftrightarrow\left[{}\begin{matrix}x=1\\left|x+1\right|+2=-\left|x-1\right|\left(\text{*}\right)\end{matrix}\right.$

Dễ thấy: $pt\left(\text{*}\right)$ vô nghiệm do $VT>0;VP\le0$

3.Dễ thấy: $VT\ge0\Rightarrow VP\ge0\Rightarrow2x-1\ge0\Rightarrow x\ge\dfrac{1}{2}$

Vậy $pt\Leftrightarrow x-2+3x-3=2x-1$

$\Leftrightarrow4x-5=2x-1$

$\Leftrightarrow2x=4\Leftrightarrow x=2$ (thỏa mãn)Câu trả lời: 1.$\left|x^2-1\right|+\left|x-1\right|=0$