Đề bài

Rút gọn biểu thức: a. \(A = \sqrt {40_{}^2 - 24_{}^2} \); b. \(B = \left( {\sqrt {12}  + 2\sqrt 3  - \sqrt {27} } \right).\sqrt 3 \); c. \(C = \frac{{\sqrt {63_{}^3 + 1} }}{{\sqrt {63_{}^2 - 62} }}\); d. \(D = \sqrt {60}  - 5\sqrt {\frac{3}{5}}  - 3\sqrt {\frac{5}{3}} \).

Lời giải chi tiết

a. \(A = \sqrt {40_{}^2 - 24_{}^2} \) \(\begin{array}{l} = \sqrt {\left( {40 - 24} \right)\left( {40 + 24} \right)} \\ = \sqrt {16.64}  = \sqrt {16} .\sqrt {64} \\ = 4.8 = 32\end{array}\) b. \(B = \left( {\sqrt {12}  + 2\sqrt 3  - \sqrt {27} } \right).\sqrt 3 \) \(\begin{array}{l} = \left( {2\sqrt 3  + 2\sqrt 3  - 3\sqrt 3 } \right).\sqrt 3 \\ = \sqrt 3.\sqrt 3 \\ = 3\end{array}\) c. \(C = \frac{{\sqrt {{{63}^3} + 1} }}{{\sqrt {{{63}^2} - 62} }}\) \(\begin{array}{l} = \frac{{\sqrt {\left( {63 + 1} \right)\left( {63_{}^2 - 63 + 1} \right)} }}{{\sqrt {63_{}^2 - 62} }}\\ = \frac{{\sqrt {64.\left( {63_{}^2 - 62} \right)} }}{{\sqrt {63_{}^2 - 62} }}\\ = \frac{{\sqrt {64} .\sqrt {63_{}^2 - 62} }}{{\sqrt {63_{}^2 - 62} }}\\ = \sqrt {64} \\ = 8\end{array}\) d. \(D = \sqrt {60}  - 5\sqrt {\frac{3}{5}}  - 3\sqrt {\frac{5}{3}} \) \(\begin{array}{l} = \sqrt {4.15}  - \sqrt {5^2.\frac{3}{5}}  - \sqrt {3^2.\frac{5}{3}}\\ = 2\sqrt {15}- \sqrt {15} - \sqrt {15}\\ = 0\end{array}\)