\(\dfrac{3}{x-2}\ge\dfrac{5}{2x-1}\)
ĐKXĐ: x ≠ 2; \(x\ne\dfrac{1}{2}\)
\(\dfrac{3}{x-2}\ge\dfrac{5}{2x-1}\)
\(\Leftrightarrow\dfrac{3}{x-2}-\dfrac{5}{2x-1}\ge0\)
\(\Leftrightarrow\dfrac{3\left(2x-1\right)-5\left(x-2\right)}{\left(x-2\right)\left(2x-1\right)}\ge0\)
\(\Leftrightarrow\dfrac{x+7}{\left(x-2\right)\left(2x-1\right)}\ge0\)
*Với: \(\dfrac{x+7}{\left(x-2\right)\left(2x-1\right)}=0\)
=> x + 7 = 0
<=> x =-7
*Với \(\dfrac{x+7}{\left(x-2\right)\left(2x-1\right)}>0\) (1)
Ta lâpj bảng xét dấu:
x |
| -7 |
| 1/2 |
| 2 |
|
X + 7 | - | 0 | + | | | + | | | + |
2x – 1 | - | | | - | 0 | + | | | + |
X - 2 | - | | | - | | | - | 0 | + |
BĐT (1) | - | 0 | + | || | - | || | + |
Từ bảng trên ta có thể thấy: \(\dfrac{x+7}{\left(x-2\right)\left(2x-1\right)}>0\) khi -7 < x < 1/2 hoăcj x > 2
Vayj:.............