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.\(\begin{array}{l}a)\,\,T = \dfrac{{ {5\sqrt x - 2} }}{{\sqrt x + 3}}\\b)\,\,x = \dfrac{1}{{11}}\end{array}\)
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.\(\begin{array}{l}a)\,\,T = \dfrac{{ {5\sqrt x - 2} }}{{\sqrt x + 3}}\\b)\,\,x = \dfrac{1}{{11}}\end{array}\)
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Hóa Học lớp 12
