Đáp án:
\(\begin{array}{l}
\left[ \begin{array}{l}
C = - 2\\
C = - 4x + 4
\end{array} \right.\\
B = \dfrac{1}{{\sqrt x + 1}}\\
B = \dfrac{{\sqrt x }}{{\sqrt x - 5}}\\
A = \dfrac{{\sqrt x }}{{\sqrt x + 3}}\\
D = \dfrac{{\left( {x + 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {2\sqrt x - 1} \right)}}\\
P = \dfrac{{\sqrt x - 4}}{{ - 15\sqrt x }}
\end{array}\)
Giải thích các bước giải:
\(\begin{array}{l}
C = \sqrt {{{\left( {2x - 3} \right)}^2}} - 2x + 1\\
= \left| {2x - 3} \right| - 2x + 1\\
\to \left[ \begin{array}{l}
C = 2x - 3 - 2x + 1\left( {DK:x \ge \dfrac{3}{2}} \right)\\
C = - 2x + 3 - 2x + 1\left( {DK:x < \dfrac{3}{2}} \right)
\end{array} \right.\\
\to \left[ \begin{array}{l}
C = - 2\\
C = - 4x + 4
\end{array} \right.\\
B = \dfrac{{\sqrt x }}{{\sqrt x \left( {\sqrt x + 1} \right)}} = \dfrac{1}{{\sqrt x + 1}}\\
B = \dfrac{{\sqrt x \left( {\sqrt x + 5} \right)}}{{\left( {\sqrt x + 5} \right)\left( {\sqrt x - 5} \right)}} = \dfrac{{\sqrt x }}{{\sqrt x - 5}}\\
A = \dfrac{{2\sqrt x \left( {\sqrt x + 3} \right) - x - 9\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{2x + 6\sqrt x - x - 9\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}}\\
= \dfrac{{x - 3\sqrt x }}{{\left( {\sqrt x + 3} \right)\left( {\sqrt x - 3} \right)}} = \dfrac{{\sqrt x }}{{\sqrt x + 3}}\\
D = \dfrac{{1 + x}}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{2\left( {\sqrt x - 1} \right) + 1}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}\\
= \dfrac{{x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x + 1} \right)}}{{2\sqrt x - 1}}\\
= \dfrac{{\left( {x + 1} \right)\left( {\sqrt x + 1} \right)}}{{\sqrt x \left( {2\sqrt x - 1} \right)}}\\
P = \dfrac{{\sqrt x - \sqrt x + 1}}{{\sqrt x \left( {\sqrt x - 1} \right)}}:\dfrac{{\left( {\sqrt x + 4} \right)\left( {\sqrt x - 4} \right) - \left( {\sqrt x + 1} \right)\left( {\sqrt x - 1} \right)}}{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 4} \right)}}\\
= \dfrac{1}{{\sqrt x \left( {\sqrt x - 1} \right)}}.\dfrac{{\left( {\sqrt x - 1} \right)\left( {\sqrt x - 4} \right)}}{{x - 16 - x + 1}}\\
= \dfrac{{\sqrt x - 4}}{{ - 15\sqrt x }}
\end{array}\)
( ý A dòng thứ A xem lại đề nha )
