Tính:
\(\frac{3}{4}.37\frac{1}{2}-\frac{3}{4}.13\frac{1}{2}\)
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TM
21 tháng 9 2017
\(P=\frac{\frac{3}{7}-\frac{3}{13}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{13}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}\)
\(=\frac{16}{35}\)
UT
4
3 tháng 12 2018
Ta có:
\(\frac{3}{4}.37\frac{1}{2}-\frac{3}{4}.13\frac{1}{2}\)
=\(\frac{3}{4}.\frac{75}{2}-\frac{3}{4}.\frac{27}{2}\)
=\(\frac{3}{4}.\left(\frac{75}{2}-\frac{27}{2}\right)\)
=\(\frac{3}{4}.24\)
=18
3 tháng 12 2018
\(\frac{3}{4}\left(37\frac{1}{2}-13\frac{1}{2}\right)\)
\(\frac{3}{4}.24=18\)
BA
18 tháng 12 2018
\(=\frac{3}{8}.37-\frac{3}{8}.13\)
\(=\frac{3}{8}\left(37-13\right)\)
\(=\frac{3}{8}.24\)
\(=9\)
5 tháng 7 2016
\(A=\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}+\frac{1.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{-1}{7}\)
\(=\frac{21}{35}-\frac{5}{35}\)
\(=\frac{16}{35}\)
5 tháng 7 2016
\(A=\frac{3.\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}{5.\left(\frac{1}{7}-\frac{1}{17}-\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{7.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(A=\frac{3}{5}+\frac{1}{7}=\frac{21}{35}+\frac{5}{35}=\frac{26}{35}\)
5 tháng 7 2019
A=\(\frac{\frac{3}{7}-\frac{3}{17}+\frac{3}{37}}{\frac{5}{7}-\frac{5}{17}+\frac{5}{37}}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{\frac{7}{5}-\frac{7}{4}+\frac{7}{3}-\frac{7}{2}}\)
\(=\frac{3\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}{5\left(\frac{1}{7}-\frac{1}{17}+\frac{1}{37}\right)}+\frac{\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}}{-7\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{4}-\frac{1}{5}\right)}\)
\(=\frac{3}{5}+\frac{1}{-7}=\frac{3}{5}-\frac{1}{7}\)
\(=\frac{21}{35}-\frac{5}{35}=\frac{16}{35}\)
Lời giải:
Ta có:
\(A=\frac{3}{4}.37\frac{1}{2}-\frac{3}{4}.13\frac{1}{2}=\frac{3}{4}.\left(37\frac{1}{2}-13\frac{1}{2}\right)=\frac{3}{4}.24=18\)
Vậy: \(\frac{3}{4}.37\frac{1}{2}-\frac{3}{4}.13\frac{1}{2}=18\)
\(\frac{3}{4}.37\frac{1}{2}\) - \(\frac{3}{4}.13\frac{1}{2}\)
=> \(\frac{3}{4}.\left(37\frac{1}{2}+13\frac{1}{2}\right)\)
=> \(\frac{3}{4}.51=\frac{153}{4}\)
dễ mà