Giải: $\frac{x-3}{4}$ – $\frac{x-2}{12}$ > $\frac{x-1}{3}$ +x 09/08/2021 Bởi Aaliyah Giải: $\frac{x-3}{4}$ – $\frac{x-2}{12}$ > $\frac{x-1}{3}$ +x
$ \dfrac{x-3}{4} – \dfrac{x-2}{12} > \dfrac{x-1}{3} + x$ $\to \dfrac{3x-9}{12} – \dfrac{x-2}{12} > \dfrac{4x-4}{12} + \dfrac{12x}{12}$ $ \to \dfrac{3x-9}{12} – \dfrac{x-2}{12} – \dfrac{4x-4}{12} – \dfrac{12x}{12} > 0 $ $\to \dfrac{3x-9 – x + 2 – 4x + 4 – 12x}{12} > 0$ $ \to -14x -3 >0$ $\to -14x > 3$ $\to x < \dfrac{-3}{14}$ Vậy $ x < \dfrac{-3}{14}$ Bình luận
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$ \dfrac{x-3}{4} – \dfrac{x-2}{12} > \dfrac{x-1}{3} + x$
$\to \dfrac{3x-9}{12} – \dfrac{x-2}{12} > \dfrac{4x-4}{12} + \dfrac{12x}{12}$
$ \to \dfrac{3x-9}{12} – \dfrac{x-2}{12} – \dfrac{4x-4}{12} – \dfrac{12x}{12} > 0 $
$\to \dfrac{3x-9 – x + 2 – 4x + 4 – 12x}{12} > 0$
$ \to -14x -3 >0$
$\to -14x > 3$
$\to x < \dfrac{-3}{14}$
Vậy $ x < \dfrac{-3}{14}$