Đáp án:
1) a=5
Giải thích các bước giải:
\(\begin{array}{l}
1)f\left( x \right) = \left\{ \begin{array}{l}
\dfrac{{{x^2} + x - 2}}{{x - 1}};x \ne 1\\
a - 2x;x = 1
\end{array} \right.\\
Xet:\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \dfrac{{{x^2} + x - 2}}{{x - 1}}\\
= \mathop {\lim }\limits_{x \to 1} \dfrac{{\left( {x + 2} \right)\left( {x - 1} \right)}}{{x - 1}} = \mathop {\lim }\limits_{x \to 1} x + 2 = 1 + 2 = 3\\
f\left( 1 \right) = a - 2x = a - 2.1 = a - 2
\end{array}\)
Để hàm số liên tục tại x=1
\(\begin{array}{l}
\to \mathop {\lim }\limits_{x \to 1} f\left( x \right) = f\left( 1 \right)\\
\to a - 2 = 3\\
\to a = 5
\end{array}\)
\(\begin{array}{l}
3)\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} \left( {{x^2} - 3x} \right) = {1^2} - 3.1 = - 2\\
\mathop {\lim }\limits_{x \to 1} f\left( x \right) = \mathop {\lim }\limits_{x \to 1} 2 = 2
\end{array}\)
