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