\(C = \frac{{x + \sqrt x – 2}}{{\sqrt x + 2}}\) (với \(x \ge 0\)) A \(C = \sqrt x + 1\) B \(C = \sqrt x – 1\) C \(C = 1 – \sqrt x \) D \(C = \sqrt x \)
by \(C = \frac{{x + \sqrt x – 2}}{{\sqrt x + 2}}\) (với \(x \ge 0\)) A \(C = \sqrt x + 1\) B \(C = …
Thẻ: Toán lớp 9
\(B = \sqrt {5 + 2\sqrt {{{\left( {1 – \sqrt 2 } \right)}^2}} } \) A \(B = \sqrt 2 – 1\) B \(B = \sqrt 2 + 1\) C \(B = 1 – \sqrt 2 \) D \(B = \sqrt 2 \)
\(B = \sqrt {5 + 2\sqrt {{{\left( {1 – \sqrt 2 } \right)}^2}} } \) A \(B = \sqrt 2 – 1\) B \(B = …
\(A = \frac{1}{{2 – \sqrt 3 }} + \frac{1}{{2 + \sqrt 3 }}\) A \(A = 1\) B \(A = 2\) C \(A = 3\) D \(A = 4\)
\(A = \frac{1}{{2 – \sqrt 3 }} + \frac{1}{{2 + \sqrt 3 }}\) A \(A = 1\) B \(A = 2\) C \(A = 3\) …
Tính: \(A = \sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}} + 1.\) \(B = \left( {2\sqrt 3 + \sqrt {20} } \right)\sqrt 3 – \sqrt {60} .\) A \(A = – \sqrt 5 \,;\,\,\,\,B = 4.\) B \(A = \sqrt 5 \,;\,\,\,\,B = 6.\) C \(A = 2\,;\,\,\,\,B = 5.\) D \(A = – 2\,;\,\,\,\,B = 4.\)
Tính: \(A = \sqrt {{{\left( {\sqrt 5 – 1} \right)}^2}} + 1.\) \(B = \left( {2\sqrt 3 + \sqrt {20} } \right)\sqrt 3 – \sqrt …
Cho biểu thức: \(P = \frac{{2\sqrt x }}{{x – 1}} – \frac{1}{{1 – \sqrt x }} + \frac{{\sqrt x }}{{\sqrt x + 1}}\) với \(x \ge 0;x \ne 1\). a) Rút gọn \(P.\) b) Tìm \(x\) để \(P = 3\). A \(\begin{array}{l}{\rm{a)}}\,\,P = \frac{{\sqrt x – 1}}{{\sqrt x + 1}}\\{\rm{b)}}\,\,x = 1\end{array}\) B \(\begin{array}{l}{\rm{a)}}\,\,P = \frac{1}{{\sqrt x – 1}}\\{\rm{b)}}\,\,x = 2\end{array}\) C \(\begin{array}{l}{\rm{a)}}\,\,P = \frac{1}{{\sqrt x + 1}}\\{\rm{b)}}\,\,x = 3\end{array}\) D \(\begin{array}{l}{\rm{a)}}\,\,P = \frac{{\sqrt x + 1}}{{\sqrt x – 1}}\\{\rm{b)}}\,\,x = 4\end{array}\)
Cho biểu thức: \(P = \frac{{2\sqrt x }}{{x – 1}} – \frac{1}{{1 – \sqrt x }} + \frac{{\sqrt x }}{{\sqrt x + 1}}\) với \(x …
Rút gọn biểu thức: \(M = \frac{{3\sqrt {75} – 12\sqrt 3 + \sqrt {12} }}{5}.\) A \(M = 1.\) B \(M = \sqrt 3 .\) C \(M = – \sqrt 3 .\) D \(M = 0.\)
Rút gọn biểu thức: \(M = \frac{{3\sqrt {75} – 12\sqrt 3 + \sqrt {12} }}{5}.\) A \(M = 1.\) B \(M = \sqrt 3 .\) …
\(\sqrt {21 + 8\sqrt 5 } + \sqrt {21 – 8\sqrt 5 } \) A \( 2\sqrt 5\) B \(-2 \sqrt 5\) C \(0\) D \(8\)
\(\sqrt {21 + 8\sqrt 5 } + \sqrt {21 – 8\sqrt 5 } \) A \( 2\sqrt 5\) B \(-2 \sqrt 5\) C \(0\) D …
\(\frac{{5\sqrt 2 – 2\sqrt 5 }}{{\sqrt 5 – \sqrt 2 }} – \frac{9}{{\sqrt {10} + 1}}\) A \(-2\) B \(2\) C \(1\) D \(-1\)
\(\frac{{5\sqrt 2 – 2\sqrt 5 }}{{\sqrt 5 – \sqrt 2 }} – \frac{9}{{\sqrt {10} + 1}}\) A \(-2\) B \(2\) C \(1\) D \(-1\) Hướng …
Rút gọn biểu thức: \(A = \frac{{3\sqrt {18} – 2\sqrt 8 }}{{\sqrt {50} }}.\) A \(A = 1.\) B \(A = \sqrt 2 .\) C \(A = 5.\) D \(A = \frac{{\sqrt 2 }}{2}.\)
Rút gọn biểu thức: \(A = \frac{{3\sqrt {18} – 2\sqrt 8 }}{{\sqrt {50} }}.\) A \(A = 1.\) B \(A = \sqrt 2 .\) C …
Rút gọn: \(\begin{array}{l}a)\,\,\,A = \left( {x – y} \right)\sqrt {\frac{3}{{y – x}}} \\b)\,\,B = 2\sqrt {3x} – \sqrt {48x} + \sqrt {108x} + \sqrt {3x} \,\,\,\,\,\left( {x \ge 0} \right)\\c)\,\,\,C = \frac{1}{{1 – 5x}}\sqrt {3{x^2}\left( {25{x^2} – 10x + 1} \right)} ,\,\,\,0 \le x \le \frac{1}{5}\\d)\,\,D = 2\sqrt {25xy} + \sqrt {225{x^3}{y^3}} – 3y\sqrt {16{x^3}y} \,\,\,\,\left( {x \ge 0,\,\,y \ge 0} \right).\end{array}\) A \(\begin{array}{l} a)\,\,A = – \sqrt {3\left( {y – x} \right)} \\ b)\,\,\,B = 5\sqrt {3x} \\ c)\,\,C = x\sqrt 3 \\ d)\,\,\,D = \sqrt {xy} \left( {10 + 3xy} \right) \end{array}\) B \(\begin{array}{l} a)\,\,A = \sqrt {3\left( {y – x} \right)} \\ b)\,\,\,B = 12\sqrt {3x} \\ c)\,\,C = x\sqrt 3 \\ d)\,\,\,D = \sqrt {xy} \left( {10 + 3xy} \right) \end{array}\) C \(\begin{array}{l} a)\,\,A = – \sqrt {3\left( {y – x} \right)} \\ b)\,\,\,B = 12\sqrt {3x} \\ c)\,\,C = x\sqrt 3 \\ d)\,\,\,D = – \sqrt {xy} \left( {10 + 3xy} \right) \end{array}\) D \(\begin{array}{l} a)\,\,A = – \sqrt {3\left( {y – x} \right)} \\ b)\,\,\,B = 5\sqrt {3x} \\ c)\,\,C = -x\sqrt 3 \\ d)\,\,\,D = \sqrt {xy} \left( {10 + 3xy} \right) \end{array}\)
Rút gọn: \(\begin{array}{l}a)\,\,\,A = \left( {x – y} \right)\sqrt {\frac{3}{{y – x}}} \\b)\,\,B = 2\sqrt {3x} – \sqrt {48x} + \sqrt {108x} + \sqrt {3x} …