\(A = 3\sqrt {32} – 6\sqrt 2 – \sqrt {50} \) A \(A = \sqrt 5 \) B \(A = \sqrt 3 \) C \(A = \sqrt 7 \) D \(A = \sqrt 2 \)
by \(A = 3\sqrt {32} – 6\sqrt 2 – \sqrt {50} \) A \(A = \sqrt 5 \) B \(A = \sqrt 3 \) C \(A …
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\(B = \sqrt {{{\left( {5 + \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {2 – \sqrt 3 } \right)}^2}} \) A \(B = 5\) B \(B = 7\) C \(B = 6\) D \(B = 8\)
\(B = \sqrt {{{\left( {5 + \sqrt 3 } \right)}^2}} + \sqrt {{{\left( {2 – \sqrt 3 } \right)}^2}} \) A \(B = 5\) B …
\(\sqrt {{{\left( {2\sqrt 5 – 5} \right)}^2}} + \sqrt {24 – 8\sqrt 5 } \) A \(\sqrt 5 \) B \(2\) C \(3\sqrt 5 \) D \(3\)
\(\sqrt {{{\left( {2\sqrt 5 – 5} \right)}^2}} + \sqrt {24 – 8\sqrt 5 } \) A \(\sqrt 5 \) B \(2\) C \(3\sqrt 5 \) …
\(4\sqrt {12} – 15\sqrt {\frac{1}{3}} – \frac{{9 – \sqrt 3 }}{{\sqrt 3 }}\) A \(1\) B \(2\) C \(3\) D \(4\)
\(4\sqrt {12} – 15\sqrt {\frac{1}{3}} – \frac{{9 – \sqrt 3 }}{{\sqrt 3 }}\) A \(1\) B \(2\) C \(3\) D \(4\) Hướng dẫn Chọn đáp …
\(\sqrt {{{\left( {2 – \sqrt 5 } \right)}^2}} – \sqrt {\frac{8}{{7 – 3\sqrt 5 }}} \). A \(- 3\) B \(- 4\) C \(- 5\) D \(- 6\)
\(\sqrt {{{\left( {2 – \sqrt 5 } \right)}^2}} – \sqrt {\frac{8}{{7 – 3\sqrt 5 }}} \). A \(- 3\) B \(- 4\) C \(- 5\) …
\(\frac{3}{{\sqrt 7 – 1}} – \frac{{\sqrt 7 – \sqrt {21} }}{{2 – 2\sqrt 3 }}\) A \(\frac{1}{2}\) B \(\frac{{\sqrt 7 }}{2}\) C \(\frac{{ – 1}}{2}\) D \(\frac{{ – \sqrt 7 }}{2}\)
\(\frac{3}{{\sqrt 7 – 1}} – \frac{{\sqrt 7 – \sqrt {21} }}{{2 – 2\sqrt 3 }}\) A \(\frac{1}{2}\) B \(\frac{{\sqrt 7 }}{2}\) C \(\frac{{ – 1}}{2}\) …
\(3\sqrt {80} – 2\sqrt {45} – \sqrt {125} \) A \(\sqrt 5 \) B \(2\sqrt 5 \) C \(3\sqrt 5 \) D \( – \sqrt 5 \)
\(3\sqrt {80} – 2\sqrt {45} – \sqrt {125} \) A \(\sqrt 5 \) B \(2\sqrt 5 \) C \(3\sqrt 5 \) D \( – \sqrt …
\(2\sqrt {48} + \frac{1}{3}\sqrt {108} – 5\sqrt 3 – 3\sqrt {27} \). A \( – 4\sqrt 3 \) B \(4\sqrt 3 \) C \(5\sqrt 3 \) D \( – 5\sqrt 3 \)
\(2\sqrt {48} + \frac{1}{3}\sqrt {108} – 5\sqrt 3 – 3\sqrt {27} \). A \( – 4\sqrt 3 \) B \(4\sqrt 3 \) C \(5\sqrt 3 …
\(\frac{{6 – \sqrt 6 }}{{\sqrt 6 – 1}} – 9\sqrt {\frac{2}{3}} – \frac{4}{{2 – \sqrt 6 }}\). A \(4\sqrt 6 \) B \(4\) C \(5\) D \(5\sqrt 6 \)
\(\frac{{6 – \sqrt 6 }}{{\sqrt 6 – 1}} – 9\sqrt {\frac{2}{3}} – \frac{4}{{2 – \sqrt 6 }}\). A \(4\sqrt 6 \) B \(4\) C \(5\) …
a) Tính chu vi tam giác ABC biết độ dài 3 cạnh là \(AB = 5\sqrt 2 \left( {cm} \right),AC = \sqrt {32} \left( {cm} \right)\), \(BC = \sqrt {98} \left( {cm} \right)\). (Không yêu cầu vẽ hình) b) Thu gọn: \(B = \sqrt {{{\left( {\sqrt 7 – 1} \right)}^2}} – \frac{6}{{\sqrt 7 – 1}}\). A \(\begin{array}{l}a)\,\,16\sqrt 2 \,\,cm\\b)\,\,B = – 2\end{array}\) B \(\begin{array}{l}a)\,\,16\,\,cm\\b)\,\,B = \sqrt 7 + 1\end{array}\) C \(\begin{array}{l}a)\,\,18\sqrt 2 \,\,cm\\b)\,\,B = \sqrt 7 – 1\end{array}\) D \(\begin{array}{l}a)\,\,18\,\,cm\\b)\,\,B = \frac{1}{{\sqrt 7 – 1}}\end{array}\)
a) Tính chu vi tam giác ABC biết độ dài 3 cạnh là \(AB = 5\sqrt 2 \left( {cm} \right),AC = \sqrt {32} \left( {cm} …