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\(6\frac{1}{2}-2\frac{1}{3}\div\frac{7}{15}\)
=\(\frac{13}{2}-\frac{7}{3}\div\frac{7}{15}\)
=\(\frac{13}{2}-5\)
=\(\frac{3}{2}\)
\(M=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{99.100}\)
\(\Rightarrow M=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}\)
\(\Rightarrow M=1-\frac{1}{100}\)
\(\Rightarrow M=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
\(b,N=\frac{1}{1.3}+\frac{1}{3.5}+\frac{1}{5.7}+...+\frac{1}{97.99}\)
\(\Rightarrow N=\frac{1}{2}.\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+..+\frac{1}{97}-\frac{1}{99}\right)\)
\(\Rightarrow N=\frac{1}{2}.\left(1-\frac{1}{99}\right)=\frac{1}{2}.\frac{98}{99}\)
\(\Rightarrow N=\frac{1.98}{2.99}=\frac{49.2}{2.99}=\frac{49}{99}\)
\(a,M=1-\frac{1}{100}=\frac{99}{100}\)
\(b=2N=\frac{2}{1x3}+\frac{2}{3x5}+\frac{2}{5x7}+...+\frac{2}{97x99}\)
\(=1-\frac{1}{99}=\frac{98}{99}\)
=>\(N=\frac{98}{99}:2=\frac{49}{99}\)
\(\frac{1}{1x2}+\frac{1}{2x3}+...+\frac{1}{99x100}=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{99}-\frac{1}{100}=1-\frac{1}{100}=\frac{99}{100}\)
a,365,04 : 23,4 x 0,01
= 15,6 x 0,01
= 0,516
b, \(\frac{15}{4}\): \(\frac{3}{2}\)- \(\frac{1}{4}\)
= \(\frac{30}{12}\)- \(\frac{1}{4}\)
= \(\frac{104}{48}\)
= \(\frac{13}{6}\)
a, 365,04 : 23,4 . 0,01
= 15,6 . 0,01
= 1560
b, \(\frac{15}{4}\div\frac{3}{2}-\frac{1}{4}\)
\(=\frac{5}{2}-\frac{1}{4}\)
\(=\frac{9}{4}\)