Đáp án:
B19: \(\dfrac{{\sqrt x + 4}}{{\sqrt x + 2}}\)
Giải thích các bước giải:
\(\begin{array}{l}
B18:\\
P = \dfrac{{x\sqrt x - 3 - 2{{\left( {\sqrt x - 3} \right)}^2} - \left( {\sqrt x + 3} \right)\left( {\sqrt x + 1} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x\sqrt x - 3 - 2\left( {x - 6\sqrt x + 9} \right) - x - 4\sqrt x - 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x\sqrt x - 3 - 2x + 12\sqrt x - 18 - x - 4\sqrt x - 3}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x\sqrt x - 3x + 8\sqrt x - 24}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x - 3} \right)\left( {x + 8} \right)}}{{\left( {\sqrt x + 1} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x + 8}}{{\sqrt x + 1}}\\
B19:\\
A = \dfrac{{\sqrt x - 3 - 5 + \left( {\sqrt x - 2} \right)\left( {\sqrt x + 2} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x - 3 - 5 + x - 4}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x + \sqrt x - 12}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\left( {\sqrt x + 4} \right)\left( {\sqrt x - 3} \right)}}{{\left( {\sqrt x + 2} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{\sqrt x + 4}}{{\sqrt x + 2}}
\end{array}\)
