Tìm x, biết: | x+ 1/3| + |x + 4/3| = 9/2 x

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1. Đáp án + Giải thích các bước giải:

   Với `∀x` có: `|x+1/3|\ge0;|x+4/3|\ge0`

   `=>|x+1/3|+|x+4/3|\ge0`

   `=>9/2x\ge0`

   `=>x\ge0`

   `=>x+1/3>0;x+4/3>0`

   `|x+1/3|+|x+4/3|=9/2x`

   `=>x+1/3+x+4/3-9/2x=0`

   `=>(x+x-9/2x)+(1/3+4/3)=0`

   `=>-5/2x+5/3=0`

   `=>-5/2x=-5/3`

   `=>x=2/3`

    

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2. Ta có: `| x+ 1/3| + |x + 4/3|≥ |x+ 1/3 + x + 4/3|`

   Dấu `=` xảy ra khi: `|x+ 1/3 + x + 4/3| = 9/2x`

   `2x + 5/3 = 9/2x`

   `2x + 5/3 – 5/3 = (9x)/2 – 5/3`

   `2x = (9x)/2 – (2(-5/3))/2`

   `2x = (9x + 2(-5/3))/2`

   `2x = (9x – 10/3)/2`

   `2.2x = 2.(9x – 10/3)/2`

   `4x = 9x – 10/3`

   `4x – 9x = 9x – 10/3 – 9x`

   `-5x = -10/3`

   `(-5x)/(-5) = (-10/3)/(-5)`

   `⇒x = 2/3`

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