Resuelve la inecuacion si x ∈![\frac{1}{4}(x-2)+ \frac{1}{3}- [x- \frac{1}{3}(2x-1)] \leq 0](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B4%7D%28x-2%29%2B+%5Cfrac%7B1%7D%7B3%7D-+%5Bx-+%5Cfrac%7B1%7D%7B3%7D%282x-1%29%5D+%5Cleq+0+++ "\frac{1}{4}(x-2)+ \frac{1}{3}- [x- \frac{1}{3}(2x-1)] \leq 0")
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1/4(x-2)+1/3-[x-1/3(2x-1)]≤0
x-2/4+1/3-[x - 2x+1/3 ] ≤0
3x-6+4/12 - [3x-2x+1/3 ] ≤0
3x-2 / 12 - [x+1/3] ≤0
3x-2/12 - x-1/3 ≤ 0
9x-6-12x-12/36≤0
-3x-18/36≤0
-3x-18≤36(0)
-3x-18≤0
-3x≤+18
x≤-18/3
x≤-6
c.s:{-6;-7;-8;-9;-10;...;-∞}
si te ayudó dale mejor respuesta y un gracias .. suerte :)