Cho hàm số $f\left( x \right) = \left\{ \begin{array}{l}\dfrac{{\sqrt {x + 4}  - 1}}{{x - 1}}\,\,\,khi\,\,\,x > 4\\3 - x\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,khi\,\,\,\,x \le 4\end{array} \right..$

Tính f (5) + f (–5).

$- 1 \le m \le 1$

$\dfrac{3}{2} \le m \le 2$

$0 \le m \le 2$

$0 \le m < \dfrac{3}{2}$

$\overrightarrow v  = \left( { - 1;2} \right)$    

$\overrightarrow v  = \left( {1; - 1} \right)$

$\overrightarrow v  = \left( {1;1} \right)$      

$\overrightarrow v  = \left( {1;0} \right)$

${M_1}\left( {2;1} \right)$.

${M_2}\left( {1;1} \right).$

${M_3}\left( {2;0} \right).$

${M_4}\left( {0; - 2} \right).$

$A\left( {2;0} \right).$

$B\left( {3;\dfrac{1}{3}} \right).$

$C\left( {1; - 1} \right).$

$D\left( { - 1; - 3} \right).$

$f\left( { - 1} \right) = 5.$

$f\left( 2 \right) = 10.$

$f\left( { - 2} \right) = 10.$

$f\left( {\dfrac{1}{5}} \right) =  - 1.$

${\rm{D}} = \mathbb{R}.$

${\rm{D}} = \left( {1; + \infty } \right).$

${\rm{D}} = \mathbb{R}\backslash \left\{ 1 \right\}.$

${\rm{D}} = \left[ {1; + \infty } \right).$