Tháng Tư 4, 2026

Thẻ: Toán lớp 9

Hỏi Đáp

Rút gọn biểu thức \(A = \sqrt {1 + \frac{1}{{{a^2}}} + \frac{1}{{{{\left( {a + 1} \right)}^2}}}} \) với \(\left( {a > 0} \right)\) A \(A = \frac{{{a^2} + a + 1}}{{a\left( {a – 1} \right)}}\) B \(A = \frac{{{a^2} + a + 1}}{{a\left( {a + 1} \right)}}\) C \(A = \frac{{{a^2} – a + 1}}{{a\left( {a – 1} \right)}}\) D \(A = \frac{{{a^2} – a – 1}}{{a\left( {a – 1} \right)}}\)

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Rút gọn biểu thức \(A = \sqrt {1 + \frac{1}{{{a^2}}} + \frac{1}{{{{\left( {a + 1} \right)}^2}}}} \) với \(\left( {a > 0} \right)\) A \(A …

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 …