by Cho biểu thức \(T = \frac{{15\sqrt x – 11}}{{x + 2\sqrt x – 3}} – \frac{{3\sqrt x – 2}}{{\sqrt x – 1}} – \frac{{2\sqrt x + 3}}{{\sqrt x + 3}}\) với điều kiện \(x \ge 0,x \ne 1\)
a) Rút gọn T b) Tìm x để \(T = \frac{1}{2}\).
A \(\begin{array}{l}
a)\,\,T = \frac{{ – \left( {5\sqrt x – 2} \right)}}{{\sqrt x + 3}}\\
b)\,\,x = \frac{1}{{11}}
\end{array}\)
B \(\begin{array}{l}
a)\,\,T = \frac{{ – \left( {5\sqrt x – 2} \right)}}{{\sqrt x + 3}}\\
b)\,\,x = \frac{1}{{121}}
\end{array}\)
C \(\begin{array}{l}
a)\,\,T = \frac{{ {5\sqrt x – 2} }}{{\sqrt x + 3}}\\
b)\,\,x = \frac{1}{{121}}
\end{array}\)
D \(\begin{array}{l}
a)\,\,T = \frac{{ {5\sqrt x – 2} }}{{\sqrt x + 3}}\\
b)\,\,x = \frac{1}{{11}}
\end{array}\)
Hướng dẫn Chọn đáp án là: B
Lời giải chi tiết:
a) Với với điều kiện \(x \ge 0,x \ne 1\) ta có :
\(\begin{array}{l}T = \frac{{15\sqrt x – 11}}{{x + 2\sqrt x – 3}} – \frac{{3\sqrt x – 2}}{{\sqrt x – 1}} – \frac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\ \Leftrightarrow T = \frac{{15\sqrt x – 11}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}} – \frac{{3\sqrt x – 2}}{{\sqrt x – 1}} – \frac{{2\sqrt x + 3}}{{\sqrt x + 3}}\\ \Leftrightarrow T = \frac{{15\sqrt x – 11 – \left( {3\sqrt x – 2} \right)\left( {\sqrt x + 3} \right) – \left( {2\sqrt x + 3} \right)\left( {\sqrt x – 1} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ \Leftrightarrow T = \frac{{15\sqrt x – 11 – 3x – 7\sqrt x + 6 – 2x – \sqrt x + 3}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}}\\ \Leftrightarrow T = \frac{{ – 5x + 7\sqrt x – 2}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}} = \frac{{ – \left( {\sqrt x – 1} \right)\left( {5\sqrt x – 2} \right)}}{{\left( {\sqrt x – 1} \right)\left( {\sqrt x + 3} \right)}} = \frac{{ – \left( {5\sqrt x – 2} \right)}}{{\sqrt x + 3}}\end{array}\)
b) \(T = \frac{{ – \left( {5\sqrt x – 2} \right)}}{{\sqrt x + 3}} = \frac{1}{2} \Leftrightarrow – 10\sqrt x + 4 = \sqrt x + 3 \Leftrightarrow 11\sqrt x = 1 \Leftrightarrow x = \frac{1}{{121}}\)
Vậy \(x = \frac{1}{{121}}\).
