Tháng Tư 4, 2026

Thẻ: Toán lớp 9

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 \(\begin{array}{l} a)\,\,3\sqrt{3}\\ b)\,\,\sqrt {a – b} + \sqrt b \end{array}\)

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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 …

Hỏi Đáp

Tính: \(\begin{array}{l} a)\,\,\frac{x}{y}\sqrt {\frac{{{x^2}}}{{{y^4}}}} \,\,\left( {x > 0,y \ne 0} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,5xy\sqrt {\frac{{25{x^2}}}{{{y^6}}}} \,\,\left( {x < 0,y > 0} \right)\\ c)\,\,a{b^2}\sqrt {\frac{3}{{{a^2}{b^4}}}} \,\,\left( {a < 0} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,d)\,\,\sqrt {\frac{{9 + 12a + 4{a^2}}}{{{b^2}}}} \,\,\left( {a \ge – \frac{3}{2},\,\,\,b < 0} \right) \end{array}\) A \(\begin{array}{l} a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, \frac{{25{x^2}}}{y}\\ c)\,\, – \sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ – b}} \end{array}\) B \(\begin{array}{l} a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, – \frac{{25{x^2}}}{y}\\ c)\,\, – \sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ – b}} \end{array}\) C \(\begin{array}{l} a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, – \frac{{25{x^2}}}{y}\\ c)\,\, \sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ – b}} \end{array}\) D \(\begin{array}{l} a)\,\,\frac{{{x^2}}}{{{y^3}}} & & & b)\,\, – \frac{{25{x^2}}}{y}\\ c)\,\, -\sqrt 3 & & d)\,\,\frac{{3 + 2a}}{{ b}} \end{array}\)

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Tính: \(\begin{array}{l} a)\,\,\frac{x}{y}\sqrt {\frac{{{x^2}}}{{{y^4}}}} \,\,\left( {x > 0,y \ne 0} \right)\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,b)\,\,5xy\sqrt {\frac{{25{x^2}}}{{{y^6}}}} \,\,\left( {x < 0,y > 0} \right)\\ c)\,\,a{b^2}\sqrt {\frac{3}{{{a^2}{b^4}}}} \,\,\left( {a < 0} …