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a) \(\left(\frac{4}{3}-\frac{4}{6}\right)+\left(\frac{4}{6}-\frac{4}{9}\right)+\left(\frac{4}{9}-\frac{4}{10}\right)+\left(\frac{4}{12}-\frac{4}{15}\right)\)
\(=\frac{4}{15}-\frac{4}{3}=\frac{-16}{15}\)
C) bạn chỉ ần bỏ các số giống nhau thôi nhé
= 1
b)
ĐẶT BIỂU THỨC TRÊN LÀ M
TA CÓ \(2M=1+\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+.....+\frac{1}{64}\)
\(\Rightarrow2M-M=1+\frac{1}{2}+\frac{1}{4}+..+\frac{1}{64}-\frac{1}{2}+\frac{1}{4}+..+\frac{1}{128}\)
\(\Rightarrow M=1+\frac{1}{28}\)
A= \(\frac{1}{2}\)+\(\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\)
2A=2(\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\))
=1+\(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)
2A-A= (\(1+\frac{1}{2}+\frac{1}{4}+...+\frac{1}{64}\)) -(\(\frac{1}{2}+\frac{1}{4}+\frac{1}{8}+...+\frac{1}{128}\))
A=1-\(\frac{1}{128}\)
A=\(\frac{127}{128}\)
\(=\frac{4}{2x4}+\frac{4}{4x6}+\frac{4}{6x8}+...+\frac{4}{18x20}\)
\(=2\left(\frac{1}{2}-\frac{1}{4}+\frac{1}{4}-\frac{1}{6}+...+\frac{1}{18}-\frac{1}{20}\right)\)
\(=2x\left(\frac{1}{2}-\frac{1}{20}\right)\\ =2x\frac{9}{20}\\ =\frac{9}{10}\)
\(1-\frac{5}{6}+\frac{3}{4}+\frac{4}{5}\)
\(=\frac{1}{6}+\frac{3}{4}+\frac{4}{5}\)
\(=\frac{103}{60}\)
=(3/2)x(4/3)x....x(100/99)
=(3x4x5x...x100)/(2x3x...x99)
=100/2
=50
có trong câu hỏi tương tự rùi mà bn
=\(\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
=\(\frac{1.2.3.4}{2.3.4.5}\)
=\(\frac{1}{5}\)
\(\left(1-\frac{1}{2}\right)x\left(1-\frac{1}{3}\right)x\left(1-\frac{1}{4}\right)x\left(1-\frac{1}{5}\right)\)
= \(\frac{1}{2}x\frac{2}{3}x\frac{3}{4}x\frac{4}{5}\)
= \(\frac{1x2x3x4}{2x3x4x5}=\frac{1}{5}\)
Gọi \(A=\frac{1}{1.4}+\frac{1}{4.7}+\frac{1}{7.10}+...+\frac{1}{22.25}\)
\(\Leftrightarrow\)\(3A=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{22.25}\)
\(\Leftrightarrow\)\(3A=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{22}-\frac{1}{25}\)
\(\Leftrightarrow\)\(3A=1-\frac{1}{25}\)
\(\Leftrightarrow\)\(3A=\frac{24}{25}\)
\(\Leftrightarrow\)\(A=\frac{24}{25}:3\)
\(\Leftrightarrow\)\(A=\frac{24}{25}.\frac{1}{3}\)
\(\Leftrightarrow\)\(A=\frac{8}{25}\)
Vậy \(A=\frac{8}{25}\)
Đặt \(C=\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}\)
\(\Rightarrow3C=\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+....+\frac{3}{22.25}\)
\(\Rightarrow3C=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{22}-\frac{1}{25}\)
\(\Rightarrow3C=1-\frac{1}{25}=\frac{24}{25}\)
\(\Rightarrow C=\frac{24}{25}:3=\frac{8}{25}\)
Vậy \(\frac{1}{1.4}+\frac{1}{4.7}+...+\frac{1}{22.25}=\frac{8}{24}\)
1/4 + 2/5 + 6/8 + 9/15 + 6/7 = ( 1/4 + 6/8 ) + ( 2/5 + 9/15 ) + 6/7
= 1 + 1 + 6/7
= 2 + 6/7
= 20/7
\(\frac{1}{4}+\frac{2}{5}+\frac{6}{8}+\frac{9}{15}+\frac{6}{7}\)
= \(\left(\frac{1}{4}+\frac{6}{8}\right)+\left(\frac{2}{5}+\frac{9}{15}\right)+\frac{6}{7}\)
= \(1+1+\frac{6}{7}\)
= 2 + \(\frac{6}{7}\)
= \(\frac{14}{7}+\frac{6}{7}=\frac{20}{7}\)
\(\frac{1}{2}:0,5-\frac{1}{4}:0,25+\frac{1}{8};0,125-\frac{1}{10}:0,1=\frac{1}{2}:\frac{1}{2}-\frac{1}{4}:\frac{1}{4}+\frac{1}{8}:\frac{1}{8}-\frac{1}{10}:\frac{1}{10}=1-1+1-1=0+0=0\)
ko
1254.1252-4/1251.1255-1
= 1254.(1251+1)-4/1251.(1254+1)-1
= 1254.1251+1254-4/1251.1254+1251-1
= 1254.1251+1250/1251.1254+1250
= 1