a) Ta có
$bx^2 +x - a \leq 0$
$<-> (x\sqrt{b})^2 + 2 . (x\sqrt{b}) . \dfrac{1}{2\sqrt{b}} + \dfrac{1}{4b} \leq a + \dfrac{1}{4b}$
$<-> \left( x \sqrt{b} + \dfrac{1}{2\sqrt{b}} \right)^2 \leq \dfrac{4ab + 1}{4b}$
Nếu $\dfrac{4ab + 1}{4b} < 0$ thì ptrinh vô nghiệm.
Ngược lại, $\dfrac{4ab + 1}{4b} \geq 0$ , ta có
$- \dfrac{4ab + 1}{4b} \leq x \sqrt{b} + \dfrac{1}{2\sqrt{b}} \leq \dfrac{4ab + 1}{4b}$
$<-> - \dfrac{4ab + 1}{4b} - \dfrac{1}{2\sqrt{b}} \leq x\sqrt{b} \leq \dfrac{4ab + 1}{4b} - \dfrac{1}{2\sqrt{b}}$
$<-> -\dfrac{4ab + 1}{4b\sqrt{b}} - \dfrac{1}{2b} \leq x \leq \dfrac{4ab + 1}{4b\sqrt{b}} - \dfrac{1}{2b}$
