Đề bài
Tính :
a] \[\sqrt {\dfrac{{{{20}^2} - {{16}^2}}}{{16}}} \]; b] \[\sqrt {\dfrac{{2\dfrac{1}{4}.1\dfrac{7}{9}}}{{3\dfrac{6}{{25}}}}} \];
c] \[\dfrac{{\sqrt {75} }}{{\sqrt 3 }}\]; d] \[\dfrac{{\sqrt {1\dfrac{{11}}{{25}}} }}{{\sqrt {2\dfrac{{14}}{{25}}} }}\].
Phương pháp giải - Xem chi tiết
+] Sử dụng công thức; \[\sqrt {{A^2}} = \left| A \right| = \left\{ \begin{array}{l}A\;\;khi\;\;A \ge 0\\ - A\;\;khi\;A < 0\end{array} \right.,\]
\[\sqrt A .\sqrt B = \sqrt {AB} ,\;\;\sqrt {\dfrac{A}{B}} = \dfrac{{\sqrt A }}{{\sqrt B }}.\]
Lời giải chi tiết
\[\begin{array}{l}a]\;\sqrt {\dfrac{{{{20}^2} - {{16}^2}}}{{16}}} = \dfrac{{\sqrt {{{20}^2} - {{16}^2}} }}{{\sqrt {16} }} \\= \dfrac{{\sqrt {\left[ {20 - 16} \right]\left[ {20 + 16} \right]} }}{4} = \dfrac{{\sqrt {4.36} }}{4} \\= \dfrac{{\sqrt 4 .\sqrt {36} }}{4} = \dfrac{{2.6}}{4} = 3.\\b]\;\sqrt {\dfrac{{2\dfrac{1}{4}.1\dfrac{7}{9}}}{{3\dfrac{6}{{25}}}}} = \dfrac{{\sqrt {\dfrac{9}{4}.\dfrac{{16}}{9}} }}{{\sqrt {\dfrac{{81}}{{25}}} }} \\= \dfrac{{\sqrt 4 }}{{\dfrac{9}{5}}} = \dfrac{{5.2}}{9} = \dfrac{{10}}{9}.\\c]\;\dfrac{{\sqrt {75} }}{{\sqrt 3 }} = \sqrt {\dfrac{{75}}{3}} = \sqrt {25} = 5.\\d]\;\dfrac{{\sqrt {1\dfrac{{11}}{{25}}} }}{{\sqrt {2\dfrac{{14}}{{25}}} }} = \dfrac{{\sqrt {\dfrac{{36}}{{25}}} }}{{\sqrt {\dfrac{{64}}{{25}}} }} = \dfrac{{\dfrac{6}{5}}}{{\dfrac{8}{5}}}\\ = \dfrac{6}{5}:\dfrac{8}{5} = \dfrac{6}{5}.\dfrac{5}{8} = \dfrac{6}{8} = \dfrac{3}{4}.\end{array}\]