Question

4, tìm x biết;

a, \(\left|x-1\right|+\left|x-3\right|=4\)

b. \(\left|2x-1\right|+\left|x+2\right|=3\)

Answer 1

a) Ta có: |x-1|+|x-3|=4(1)

Trường hợp 1: x<1

(1)\(\Leftrightarrow1-x+3-x=4\)

\(\Leftrightarrow4-2x=4\)

\(\Leftrightarrow2x=0\)

hay x=0(thỏa mãn)

Trường hợp 2: \(1\le x< 3\)

(1)\(\Leftrightarrow x-1+3-x=4\)

\(\Leftrightarrow2=4\)(vô lý)

Trường hợp 3: x≥3

(1)\(\Leftrightarrow x-1+x-3=4\)

\(\Leftrightarrow2x-4=4\)

\(\Leftrightarrow2x=8\)

hay x=4(thỏa mãn)

Vậy: x∈{0;4}

b) Ta có: |2x-1|+|x+2|=3(2)

Trường hợp 1: x<-2

(2)\(\Leftrightarrow-2-x-2x+1=3\)

\(\Leftrightarrow-3x-1=3\)

\(\Leftrightarrow-3x=4\)

hay \(x=\frac{-4}{3}\)(loại)

Trường hợp 2: \(-2\le x< \frac{1}{2}\)

(2)\(\Leftrightarrow x+2-2x+1=3\)

\(\Leftrightarrow-x+3=3\)

\(\Leftrightarrow-x=0\)

hay x=0(thỏa mãn)

Trường hợp 3: \(x\ge\frac{1}{2}\)

(2)\(\Leftrightarrow2x-1+x+2=3\)

\(\Leftrightarrow3x+1=3\)

\(\Leftrightarrow3x=2\)

hay \(x=\frac{2}{3}\)(thỏa mãn)

Vậy: \(x\in\left\{0;\frac{2}{3}\right\}\)