Question

bài 1

Tìm GTNN của |x-3|+|x-20|

bài 2

Có \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)

CM\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\)

Answer 1

\(\dfrac{x}{2}=\dfrac{y}{3}=\dfrac{z}{4}\Rightarrow\left\{{}\begin{matrix}x=\dfrac{2y}{3}\\x=\dfrac{z}{2}\\y=\dfrac{3z}{4}\end{matrix}\right.\)

ta có \(\dfrac{3x-2y}{4}=\dfrac{2z-4x}{3}=\dfrac{4y-3z}{2}\)

\(\Leftrightarrow\dfrac{3\dfrac{2y}{3}-2y}{4}=\dfrac{2z-4\dfrac{z}{2}}{3}=\dfrac{4\dfrac{3z}{4}-3z}{2}\)

\(\Leftrightarrow0=0=0\)( luôn đúng)

suy ra Đpcm

Answer 2

Bài 1\(\left|x-3\right|+\left|20-x\right|\ge\left|x-3+20-x\right|=17\)

dấu "=" xảy ra khi x-3=20-x \(\Rightarrow x=\dfrac{23}{2}\)