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Thay vào:
|x−1|+1−2[|x−2|+2]=−3|x−1|+1−2[|x−2|+2]=−3
⇔|x−1|−2|x−2|=−3−1+4=0⇔⇔|x−1|−2|x−2|=−3−1+4=0⇔
|x−1|−2|x−2|=0|x−1|−2|x−2|=0(1)
Chia khoảng ⎧⎩⎨⎪⎪x<1|x−1|=1−x|x−2|=2−x{x<1|x−1|=1−x|x−2|=2−x⇒(1)⇔1−x−4+2x=0⇒x=3>1⇒(1)⇔1−x−4+2x=0⇒x=3>1(LOẠI)
⎧⎩⎨⎪⎪1≤x<2|x−1|=x−1|x−2|=2−x{1≤x<2|x−1|=x−1|x−2|=2−x⇒x−1−4+2x=0⇒x=53<2⇒x−1−4+2x=0⇒x=53<2(NHẬN)
⎧⎩⎨⎪⎪x≥2|x−1|=x−1|x−2|=x−2{x≥2|x−1|=x−1|x−2|=x−2⇒x−1+4−2x=0⇒x=3>2⇒x−1+4−2x=0⇒x=3>2(nhận)
Kết luận: ⎡⎣x=53x=3
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(f\left(x\right)-2\cdot g\left(x\right)=-3\)
\(\Leftrightarrow\left|x-1\right|+1-2\left|x-2\right|-4=-3\)
\(\Leftrightarrow\left|x-1\right|-2\left|x-2\right|=0\)
=>|x-1|=|2x-4|
=>2x-4=x-1 hoặc 2x-4=-x+1
=>x=3 hoặc x=5/3
b: f(x)=g(f(2))
\(\Leftrightarrow\left|x-1\right|+1=g\left(\left|1-2\right|+1\right)=g\left(2\right)\)
=>|x-1|=1
=>x=2 hoặc x=0
![](https://rs.olm.vn/images/avt/0.png?1311)
Thay vào:
\(\left|x-1\right|+1-2\left[\left|x-2\right|+2\right]=-3\)
\(\Leftrightarrow\left|x-1\right|-2\left|x-2\right|=-3-1+4=0\Leftrightarrow\)
\(\left|x-1\right|-2\left|x-2\right|=0\)(1)
Chia khoảng \(\left\{\begin{matrix}x< 1\\\left|x-1\right|=1-x\\\left|x-2\right|=2-x\end{matrix}\right.\)\(\Rightarrow\left(1\right)\Leftrightarrow1-x-4+2x=0\Rightarrow x=3>1\)(LOẠI)
\(\left\{\begin{matrix}1\le x< 2\\\left|x-1\right|=x-1\\\left|x-2\right|=2-x\end{matrix}\right.\)\(\Rightarrow x-1-4+2x=0\Rightarrow x=\dfrac{5}{3}< 2\)(NHẬN)
\(\left\{\begin{matrix}x\ge2\\\left|x-1\right|=x-1\\\left|x-2\right|=x-2\end{matrix}\right.\)\(\Rightarrow x-1+4-2x=0\Rightarrow x=3>2\)(nhận)
Kết luận: \(\left[\begin{matrix}x=\dfrac{5}{3}\\x=3\end{matrix}\right.\)