Câu hỏi
a) Tính: \(A=\left( \frac{2}{3-\sqrt{5}}+\frac{1}{2-\sqrt{5}}-\frac{10}{2\sqrt{5}} \right).\left( 1-3\sqrt{5} \right)\)
b) Rút gọn biểu thức sau: \(P=\frac{\sqrt{x}}{2\sqrt{x}-3}+\frac{\sqrt{x}-2}{2\sqrt{x}+3}+\frac{15-4\sqrt{x}}{9-4x}\) (với \(x\ge 0{{;}^{{}}}^{{}}x\ne \frac{9}{4}\))
Lời giải chi tiết:
a) Ta có:
\(\begin{array}{l}A = \left( {\frac{2}{{3 - \sqrt 5 }} + \frac{1}{{2 - \sqrt 5 }} - \frac{{10}}{{2\sqrt 5 }}} \right).\left( {1 - 3\sqrt 5 } \right)\\ = \left( {\frac{{2\left( {3 + \sqrt 5 } \right)}}{{\left( {3 - \sqrt 5 } \right)\left( {3 + \sqrt 5 } \right)}} + \frac{{1.\left( {2 + \sqrt 5 } \right)}}{{\left( {2 + \sqrt 5 } \right)\left( {2 - \sqrt 5 } \right)}} - \frac{5}{{\sqrt 5 }}} \right).\left( {1 - 3\sqrt 5 } \right)\\ = \left( {\frac{{2\left( {3 + \sqrt 5 } \right)}}{4} + \frac{{2 + \sqrt 5 }}{{ - 1}} - \sqrt 5 } \right).\left( {1 - 3\sqrt 5 } \right)\\ = \left( {\frac{{3 + \sqrt 5 }}{2} - \frac{{2 + \sqrt 5 }}{1} - \sqrt 5 } \right).\left( {1 - 3\sqrt 5 } \right)\\ = \left( {\frac{{3 + \sqrt 5 - 2\left( {2 + \sqrt 5 } \right) - 2\sqrt 5 }}{2}} \right).\left( {1 - 3\sqrt 5 } \right)\\= \left( {\frac{{3 + \sqrt 5 - 4 - 2\sqrt 5 - 2\sqrt 5 }}{2}} \right).\left( {1 - 3\sqrt 5 } \right)\\ = - \left( {\frac{{ - 1 - 3\sqrt 5 }}{2}} \right).\left( {3\sqrt 5 - 1} \right)\\ = \left( {\frac{{3\sqrt 5 + 1}}{2}} \right).\left( {3\sqrt 5 - 1} \right)\\ = \frac{{\left( {3\sqrt 5 + 1} \right)\left( {3\sqrt 5 - 1} \right)}}{2}\\ = \frac{{44}}{2} = 22\end{array}\)
b) Ta có:
\(\begin{array}{l}P = \frac{{\sqrt x }}{{2\sqrt x - 3}} + \frac{{\sqrt x - 2}}{{2\sqrt x + 3}} + \frac{{15 - 4\sqrt x }}{{9 - 4x}}\\ = \frac{{\sqrt x }}{{2\sqrt x - 3}} + \frac{{\sqrt x - 2}}{{2\sqrt x + 3}} - \frac{{15 - 4\sqrt x }}{{4x - 9}}\\ = \frac{{\sqrt x }}{{2\sqrt x - 3}} + \frac{{\sqrt x - 2}}{{2\sqrt x + 3}} - \frac{{15 - 4\sqrt x }}{{\left( {2\sqrt x - 3} \right).\left( {2\sqrt x + 3} \right)}}\\ = \frac{{\sqrt x .\left( {2\sqrt x + 3} \right) + \left( {\sqrt x - 2} \right).\left( {2\sqrt x - 3} \right) - \left( {15 - 4\sqrt x } \right)}}{{\left( {2\sqrt x - 3} \right).\left( {2\sqrt x + 3} \right)}}\\ = \frac{{2x + 3\sqrt x + 2x - 3\sqrt x - 4\sqrt x + 6 - 15 + 4\sqrt x }}{{\left( {2\sqrt x - 3} \right).\left( {2\sqrt x + 3} \right)}}\\ = \frac{{4x - 9}}{{4x - 9}}\\ = 1\end{array}\)
Câu hỏi trước
