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1/1+2+1/1+2+3+1/1+2+3+4+1/1+2+3+4+5
=1/3+1/6+1/10+1/15
lấy mẫu số chung là 30,ta có:
10/30+5/6+3/10+2/30
=20/30=2/3
\(M=\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+\frac{1}{1+2+3+4+5}\)
\(M=\frac{1}{\left(1+2\right).2:2}+\frac{1}{\left(1+3\right).3:2}+\frac{1}{\left(1+4\right).4:2}+\frac{1}{\left(1+5\right).5:2}\)
\(M=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}\)
\(M=2.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\right)\)
\(M=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\right)\)
\(M=2.\left(\frac{1}{2}-\frac{1}{6}\right)\)
\(M=2.\frac{1}{2}-2.\frac{1}{6}\)
\(M=1-\frac{1}{3}=\frac{2}{3}\)
a,11/5x3/2-3/1/2x3/5
=11/2x3/5-7/2-3/5
=3/5x(11/2-7/2)
=3/5x2
=6/5
b,(1/2-1/3+1/4-1/5):(1/4-1/6)
=(1/6+1/20):1/12
=13/60:1/2
=13/30
Ta có:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)........\left(1-\frac{1}{2017}\right).\left(1-\frac{1}{2018}\right)\)
\(\Rightarrow B=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.......\frac{2016}{2017}.\frac{2017}{2018}\)
Đởn giản hết sẽ còn là:
\(\Rightarrow B=\frac{1}{2018}\)
=17/6+1/2×13/3-9/4:3/5
=17/6+13/6-9/4×5/3
=17/6+13/6-15/4
=68+52-90/24
=30/24
=5/4
\(2\frac{5}{6}+0,5\times4\frac{1}{3}-2\frac{1}{4}:\frac{3}{5}\)
\(=\frac{17}{6}+\frac{1}{2}\times\frac{13}{3}-\frac{9}{4}:\frac{3}{5}\)
\(=\frac{17}{6}+\frac{13}{6}-\frac{9}{4}\times\frac{5}{3}\)
\(=5-\frac{9\times5}{4\times3}\)
\(=5-\frac{3\times5}{4\times1}\)
\(=5-\frac{15}{4}=\frac{20-15}{4}=\frac{5}{4}\)
Bài 2:
\(B=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right).......\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}....\frac{2003}{2004}\)
\(=\frac{1}{2004}\)