Tháng Năm 23, 2026

Tác giả: admin

Hỏi Đáp

Tìm \(x\) biết: \(a)\,\,\frac{{\sqrt {2x – 1} }}{{\sqrt {x – 1} }} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,\frac{{\sqrt {{x^2} – 4} }}{{\sqrt {x – 2} }} = 3\) A \(\begin{array}{l} a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\ b)\,\,S = \left\{- 7 \right\}. \end{array}\) B \(\begin{array}{l} a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\ b)\,\,S = \left\{ 7 \right\}. \end{array}\) C \(\begin{array}{l} a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\ b)\,\,S = \left\{ 2 \right\}. \end{array}\) D \(\begin{array}{l} a)\,\,\,S = \left\{ {\frac{3}{2}} \right\}.\\ b)\,\,S = \left\{ 2;\,7 \right\}. \end{array}\)

adminby admin

Tìm \(x\) biết: \(a)\,\,\frac{{\sqrt {2x – 1} }}{{\sqrt {x – 1} }} = 2\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,\frac{{\sqrt {{x^2} – 4} }}{{\sqrt {x – 2} }} = 3\) A …

Hỏi Đáp

Rút gọn biểu thức \(\begin{array}{l}a)\,\,\left( {5\sqrt {48} + 4\sqrt {27} – 2\sqrt {12} } \right):\sqrt 3 \\b)\,\,\left( {\sqrt {{a^2} – {b^2}} + \sqrt {\left( {a + b} \right).b} } \right):\sqrt {a + b} \,\,\,\left( {a > b > 0} \right).\end{array}\) A \(\begin{array}{l} a)\,\,\sqrt{3}\\ b)\,\,\sqrt {a – b} + \sqrt b \end{array}\) B \(\begin{array}{l} a)\,\,28\\ b)\,\,\sqrt {a – b} + \sqrt b \end{array}\) C \(\begin{array}{l} a)\,\,28\\ b)\,\,\sqrt {a + b} + \sqrt b \end{array}\) D \(\begin{array}{l} a)\,\,3\sqrt{3}\\ b)\,\,\sqrt {a – b} + \sqrt b \end{array}\)

adminby admin

Rút gọn biểu thức \(\begin{array}{l}a)\,\,\left( {5\sqrt {48} + 4\sqrt {27} – 2\sqrt {12} } \right):\sqrt 3 \\b)\,\,\left( {\sqrt {{a^2} – {b^2}} + \sqrt {\left( {a …