Question

Tìm x: \(\left|x+\dfrac{1}{3}\right|+\left|x+\dfrac{2}{3}\right|+\left|x+\dfrac{2}{5}\right|+\left|x+\dfrac{3}{2}\right|=33x\)

Answer 1

`|x+1/3|+|x+2/3|+|x+2/5|+|x+3/2|=33x`

`@TH1: x >= -1/3`

  `=>x+1/3+x+2/3+x+2/5+x+3/2=33x`

 `=>29x=29/10`

 `=>x=1/10` (t/m)

`@TH2: -2/3 <= x < -1/3`

 `=>-x-1/3+x+2/3+x+2/5+x+3/2=33x`

 `=>31x=67/30`

 `=>x=67/930` (ko t/m)

`@TH3:-2/5 <= x < -2/3`

  `=>-x-1/3-x-2/3+x+2/5+x+3/2=33x`

  `=>33x=9/10`

 `=>x=3/110` (ko t/m)

`@TH4:-3/2 <= x < -2/5`

  `=>-x-1/3-x-2/3-x-2/5+x+3/2=33x`

  `=>35x=1/10`

  `=>x=1/350` (ko t/m)

`@TH5: x < -3/2`

  `=>-x-1/3-x-2/3-x-2/5-x-3/2=33x`

  `=>37x=-29/10`

  `=>x=-29/370` (ko t/m)

Answer 2

có VT \(\ge\) 0 với mọi x

=>VP:33x\(\ge\) 0 \(\Rightarrow\) x\(\ge\)0

\(\Rightarrow\) |x+1/3|\(\ge\)0;|x+2/3|\(\ge\) 0;|x+2/5|\(\ge\) 0;|x+3/2|\(\ge\) 0

=> (x+1/3)+(x+2/3)+(x+2/5)+(x+3/2)=33x

=>(x+x+x+x)+(1/3+2/3+2/5+3/2)=33x

=>4x+29/10=33x

=>  29/10=33x-4x

=>29/10=29x

=>x=29/10:29

=>x=1/10