Đáp án:
$\begin{array}{l}
1)a)Q = \left( {\dfrac{{\sqrt x }}{{1 - \sqrt x }} + \dfrac{{\sqrt x }}{{1 + \sqrt x }}} \right) + \dfrac{{3 - \sqrt x }}{{x - 1}}\\
= \dfrac{{\sqrt x \left( {1 + \sqrt x } \right) + \sqrt x \left( {1 - \sqrt x } \right)}}{{\left( {1 - \sqrt x } \right)\left( {1 + \sqrt x } \right)}} + \dfrac{{3 - \sqrt x }}{{x - 1}}\\
= \dfrac{{\sqrt x + x + \sqrt x - x}}{{1 - x}} + \dfrac{{\sqrt x - 3}}{{1 - x}}\\
= \dfrac{{2\sqrt x + \sqrt x - 3}}{{1 - x}}\\
= \dfrac{{3\sqrt x - 3}}{{1 - x}}\\
= \dfrac{{ - 3}}{{\sqrt x + 1}}\\
b)Q = - 1\\
\Rightarrow \dfrac{{ - 3}}{{\sqrt x + 1}} = - 1\\
\Rightarrow \sqrt x + 1 = 3\\
\Rightarrow \sqrt x = 2\\
\Rightarrow x = 4\left( {tmdk} \right)\\
2)a)\left( {{d_1}} \right):y = 2x - 1\\
+ Cho:x = 0 \Rightarrow y = - 1\\
+ Cho:x = 1 \Rightarrow y = 1\\
\Rightarrow \left( {0; - 1} \right);\left( {1;1} \right) \in \left( {{d_1}} \right)\\
\left( {{d_2}} \right):y = - \dfrac{1}{2}x + 4\\
+ Cho:x = 0 \Rightarrow y = 4\\
+ Cho:x = 2 \Rightarrow y = 3\\
b)2x - 1 = - \dfrac{1}{2}x + 4\\
\Rightarrow 2x + \dfrac{1}{2}x = 5\\
\Rightarrow \dfrac{5}{2}x = 5\\
\Rightarrow x = 2\\
\Rightarrow y = 2x - 1 = 3\\
\Rightarrow \left( {{d_1}} \right) \cap \left( {{d_2}} \right) = A\left( {2;3} \right)\\
c)B\left( {0; - 1} \right);C\left( {0;4} \right)\\
\Rightarrow BC = 5\\
\Rightarrow {h_A} = 2\\
\Rightarrow {S_{ABC}} = \dfrac{1}{2}.2.5 = 5
\end{array}$
