Đáp án:
m=4
Giải thích các bước giải:
\(\begin{array}{l}
A = \dfrac{{2\sqrt x - 1}}{{x + 2}}\\
B = \dfrac{{3\sqrt x }}{{2\sqrt x + 1}} - \dfrac{4}{{1 - 2\sqrt x }} - \dfrac{{4x + 2\sqrt x + 3}}{{4x - 1}}\\
= \dfrac{{3\sqrt x \left( {2\sqrt x - 1} \right) + 4\left( {2\sqrt x + 1} \right) - 4x - 2\sqrt x - 3}}{{\left( {2\sqrt x + 1} \right)\left( {2\sqrt x - 1} \right)}}\\
= \dfrac{{6x - 3\sqrt x + 8\sqrt x + 4 - 4x - 2\sqrt x - 3}}{{\left( {2\sqrt x + 1} \right)\left( {2\sqrt x - 1} \right)}}\\
= \dfrac{{2x + 3\sqrt x + 1}}{{\left( {2\sqrt x + 1} \right)\left( {2\sqrt x - 1} \right)}}\\
= \dfrac{{\left( {2\sqrt x + 1} \right)\left( {\sqrt x + 1} \right)}}{{\left( {2\sqrt x + 1} \right)\left( {2\sqrt x - 1} \right)}}\\
= \dfrac{{\sqrt x + 1}}{{2\sqrt x - 1}}\\
Có:\left( {AB - 1} \right)\left( {x + 2} \right) = m\left( {1 - \sqrt x } \right) + 3\sqrt x - 4\\
\to \left( {\dfrac{{2\sqrt x - 1}}{{x + 2}}.\dfrac{{\sqrt x + 1}}{{2\sqrt x - 1}} - 1} \right)\left( {x + 2} \right) = m\left( {1 - \sqrt x } \right) + 3\sqrt x - 4\\
\to \left( {\dfrac{{\sqrt x + 1 - x - 2}}{{x + 2}}} \right)\left( {x + 2} \right) = m\left( {1 - \sqrt x } \right) + 3\sqrt x - 4\\
\to - x + \sqrt x - 1 = m - m\sqrt x + 3\sqrt x - 4\\
\to x + 2\sqrt x - 3 + m - m\sqrt x = 0\\
\to x + \left( {2 - m} \right)\sqrt x + m - 3 = 0\left( 1 \right)
\end{array}\)
Yêu cầu đề bài ⇔ Phương trình (1) có nghiệm kép
\(\begin{array}{l}
\to 4 - 4m + {m^2} - 4m + 12 = 0\\
\to {m^2} - 8m + 16 = 0\\
\to {\left( {m - 4} \right)^2} = 0\\
\to m = 4
\end{array}\)
