Question

Tìm số nguyên x thoả mãn cả hai bpt :

a. \(\frac{3x-2}{5}\ge\frac{x}{2}+0,8\) và \(1-\frac{2x-5}{6}>\frac{3-x}{4}\)

b. 2(3x-4) < 3(4x-3) +16 và 4(1+x)<3x+5

Answer 1

a)\(\frac{3x-2}{5}\ge\frac{x}{2}+0,8\) va \(1-\frac{2x-5}{6}>\frac{3-x}{4}\)

 \(\cdot\frac{3x-2}{5}\ge\frac{x}{2}+0,8\)

  \(=\frac{2\left(3x-2\right)}{10}\ge\frac{5x}{10}+\frac{8}{10}\)

   \(\Rightarrow2\left(3x-2\right)\ge5x+8\)

   \(=6x-4\ge5x+8\)

   \(=6x-5x\ge8+4\)

    \(x\ge12\)(1)

\(\cdot1-\frac{2x-5}{6}>\frac{3-x}{4}\)

 \(=\frac{12}{12}-\frac{2\left(2x-5\right)}{12}>\frac{3\left(3-x\right)}{12}\)

  \(\Rightarrow12-2\left(2x-5\right)>3\left(3-x\right)\)

  \(=12-4x+10>9-3x\)

  \(=-4x+3x>9-12-10\)

   \(=-x>-13\)

    \(=x< 13\) (2)

Từ (1) và (2) => \(13>x\ge12\)=> x=12