Câu hỏi:
Rút gọn biểu thức
\(\begin{array}{l}a)\,\,\left( {5\sqrt {48} + 4\sqrt {27} - 2\sqrt {12} } \right):\sqrt 3 \\b)\,\,\left( {\sqrt {{a^2} - {b^2}} + \sqrt {\left( {a + b} \right).b} } \right):\sqrt {a + b} \,\,\,\left( {a > b > 0} \right).\end{array}\)
Phương pháp giải:
Lời giải chi tiết:
\(\begin{array}{l}a)\,\,\left( {5\sqrt {48} + 4\sqrt {27} - 2\sqrt {12} } \right):\sqrt 3 = \frac{{5\sqrt {48} + 4\sqrt {27} - 2\sqrt {12} }}{{\sqrt 3 }}\\ = \frac{{5\sqrt {{4^2}.3} + 4\sqrt {{3^2}.3} - 2\sqrt {{2^2}.3} }}{{\sqrt 3 }} = \frac{{5.4\sqrt 3 + 4.3\sqrt 3 - 2.2\sqrt 3 }}{{\sqrt 3 }}\\ = \frac{{20\sqrt 3 + 12\sqrt 3 - 4\sqrt 3 }}{{\sqrt 3 }} = \frac{{28\sqrt 3 }}{{\sqrt 3 }} = 28.\\b)\,\,\left( {\sqrt {{a^2} - {b^2}} + \sqrt {\left( {a + b} \right)b} } \right):\sqrt {a + b} \,\,\,\,\left( {a > b > 0} \right)\\ = \frac{{\sqrt {{a^2} - {b^2}} + \sqrt {\left( {a + b} \right).b} }}{{\sqrt {a + b} }} = \frac{{\sqrt {\left( {a + b} \right)\left( {a - b} \right)} + \sqrt b .\sqrt {a + b} }}{{\sqrt {a + b} }}\\ = \frac{{\sqrt {a + b} .\sqrt {a - b} + \sqrt b .\sqrt {a + b} }}{{\sqrt {a + b} }} = \frac{{\sqrt {a + b} \left( {\sqrt {a - b} + \sqrt b } \right)}}{{\sqrt {a + b} }} = \sqrt {a - b} + \sqrt b .\end{array}\)