Cho hàm số \(y = f\left( x \right) = \left\{ \begin{array}{l}\frac{{\sqrt[3]{{ax + 1}} - \sqrt {1 - bx} }}{x}\,\,\,khi\,\,x \ne 0\\3a - 5b - 1\,\,\,\,khi\,\,x = 0\end{array} \right...

* Hướng dẫn giải

TXĐ: \(D = \mathbb{R},\,\,x = 0 \in D\).

Ta có:

\(\begin{array}{l}\mathop {\lim }\limits_{x \to 0} f\left( x \right) = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{ax + 1}} - \sqrt {1 - bx} }}{x}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\sqrt[3]{{ax + 1}} - 1}}{x} + \mathop {\lim }\limits_{x \to 0} \frac{{1 - \sqrt {1 - bx} }}{x}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{\left( {\sqrt[3]{{ax + 1}} - 1} \right)\left( {{{\sqrt[3]{{ax + 1}}}^2} + \sqrt[3]{{ax + 1}} + 1} \right)}}{{x\left( {{{\sqrt[3]{{ax + 1}}}^2} + \sqrt[3]{{ax + 1}} + 1} \right)}}\\ + \mathop {\lim }\limits_{x \to 0} \frac{{\left( {1 - \sqrt {1 - bx} } \right)\left( {1 + \sqrt {1 - bx} } \right)}}{{x\left( {1 + \sqrt {1 - bx} } \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{{ax + 1 - 1}}{{x\left( {{{\sqrt[3]{{ax + 1}}}^2} + \sqrt[3]{{ax + 1}} + 1} \right)}}\\ + \mathop {\lim }\limits_{x \to 0} \frac{{1 - 1 + bx}}{{x\left( {1 + \sqrt {1 - bx} } \right)}}\\ = \mathop {\lim }\limits_{x \to 0} \frac{a}{{{{\sqrt[3]{{ax + 1}}}^2} + \sqrt[3]{{ax + 1}} + 1}}\\ + \mathop {\lim }\limits_{x \to 0} \frac{b}{{1 + \sqrt {1 - bx} }}\\ = \frac{a}{{1 + 1 + 1}} + \frac{b}{{1 + 1}}\\ = \frac{a}{3} + \frac{b}{2}\end{array}\)

\(f\left( 0 \right) = 3a - 5b - 1\).

Để hàm số liên tục tại \(x = 0\) thì \(\mathop {\lim }\limits_{x \to 0} f\left( x \right) = f\left( 0 \right)\)

\( \Leftrightarrow \frac{a}{3} + \frac{b}{2} = 3a - 5b - 1\) \( \Leftrightarrow \frac{8}{3}a - \frac{{11}}{2}b = 1\)\( \Leftrightarrow 16a - 33b = 6\)