Question

a)\(\frac{3-2x}{5}\)>\(\frac{2-x}{3}\)

b)\(\frac{x-2}{6}\)-\(\frac{x-1}{3}\)≤\(\frac{x}{2}\)

c)\(\frac{x+1}{3}\)>\(\frac{2x-1}{6}\)-2

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Answer 1

c) \(\frac{x+1}{3}>\frac{2x-1}{6}-2\)

⇔\(\frac{2\left(x+1\right)}{6}>\frac{2x-1-12}{6}\)

⇔2x + 2 > 2x - 13

⇔0x > -15

Vậy S=Φ

Answer 2

a) \(\frac{3-2x}{5}>\frac{2-x}{3}\)

⇔\(\frac{3\left(3-2x\right)}{15}>\frac{5\left(2-x\right)}{15}\)

⇔9 - 6x > 10 - 5x

⇔-x > 1

⇔x < -1

Vậy S={x | x < -1}

Answer 3

b) \(\frac{x-2}{6}-\frac{x-1}{3}\le\frac{x}{2}\)

⇔\(\frac{x-2-2\left(x-1\right)}{6}\le\frac{3x}{6}\)

⇔x - 2 - 2x +2 ≤ 3x

⇔-4x ≤ 0

⇔x ≥ 0

Vậy S={x | x ≥ 0}