Question

giải phương trình

\(\dfrac{x+2}{4}\)<2+\(\dfrac{x-1}{3}\)

Answer 1

\(\dfrac{x+2}{4}< \dfrac{5+x}{3}\Leftrightarrow\dfrac{3x+6}{12}< \dfrac{20+4x}{12}\Leftrightarrow\dfrac{3x+6-20-4x}{12}< 0\Leftrightarrow-x-14< 0\Leftrightarrow-\left(x+14\right)< 0\Leftrightarrow x+14>0\Leftrightarrow x>-14\)

Answer 2

<=> \(\dfrac{3\left(x+2\right)}{12}\)\(< \)\(\dfrac{24}{12}\)+\(\dfrac{4\left(x-1\right)}{12}\)

<=> 3x + 6 < 24 + 4x - 4

<=> 3x - 4x < -6 + 24 - 4

<=> -x < 14 <=> x > -14

Vây S = { x / x > -14 } đơn giản :)