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lấy (1/3 + 1/15 +1/10 + 1/21 ) + (1/36 + 1/28 + 1/6) + (1/45 + 1/55)
= (4/50 + 3/70) + 2/100
= 7/120 + 2/100
= 9/220
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{9.10}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+..+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=2.\frac{2}{5}=\frac{4}{5}\)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+\frac{1}{15}+...+\frac{1}{45}\)
\(=\frac{2}{6}+\frac{2}{12}+\frac{2}{20}+\frac{2}{30}+...+\frac{2}{90}\)
\(=\frac{2}{2.3}+\frac{2}{3.4}+\frac{2}{4.5}+\frac{2}{5.6}+...+\frac{2}{9.10}\)
\(=2.\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{9}-\frac{1}{10}\right)\)
\(=2.\left(\frac{1}{2}-\frac{1}{10}\right)\)
\(=\frac{4}{5}\)
\(A=\frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{2004\cdot2005}+\frac{1}{2005\cdot2006}\)
\(A=\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2004}-\frac{1}{2005}+\frac{1}{2005}-\frac{1}{2006}\)
\(A=1-\frac{1}{2006}=\frac{2005}{2006}\)
\(A=\frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2005.2006}\)
\(\Rightarrow A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{2005}-\frac{1}{2006}\)
\(\Rightarrow A=1-\frac{1}{2006}\)
\(\Rightarrow A=\frac{2005}{2006}\)
= 2/2 + 2/6 + 2/12+...+2/90 = 2(1/2 +1/6 + 1/12 + ...+ 1/90) = 2(1/1.2 + 1/2.3 + 1/3.4 + ... + 1/9.10) = 2(1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + ... + 1/9 - 1/10) = 2(1 - 1/10) = 2 . 9/10 = 9/5
1/3 + 1/15 + 1/35+ 1/63 +...... + 1/195
= 1/3 + 1/3x5 + 1/5 x7 + 1/7x9 + ....+1/13x15
= 1/3+1/3-1/5+1/5-1/7+1/7-1/9+....+1/13-1/15 ( vì +- nên rút gọn )
= 1/3+1/3-1/15
=3/5
=1/1.3+1/3.5+1/5.7+...+1/13.15
=1/2.2(1/1.3+1/3.5+1/5.7+...+1/13.15)
=1/2(2/1.3+2/3.5+2/5.7+...+2/13.15)
=1/2(1-1/3+1/3-1/5+1/5-1/7+...+1/13-1/15)
=1/2[(1-1/15)+(1/3-1/3)+(1/5-1/5)+...+(1/13-1/15)]
=1/2[(1-1/15)+0+...+0=1/2(1-1/15)=1/2.14/15=14/30=7/15
=(1+1+1+1+1+1+1+1)+(1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45)
Đặt A = 1/3+1/6+1/10+1/15+1/21+1/28+1/36+1/45
Ta có:
A x 1/2= 1/6+1/12+1/20+1/30+1/42+1/56+1/72+1/90
1/6=1/2x3=1/2-1/3
1/12=1/3x4=1/3-1/4
……………………
1/90=1/9x10=1/9-1/10
A x 1/2=1/2-1/3+1/3-1/4+1/4-1/5+…+1/9-1/10
A x 1/2=1/2-1/10=4/10
A=4/10:1/2=4/5
Vậy 4/3+7/6+11/10+16/15+22/21+29/28+37/36+46/45=1+1+1+1+1+1+1+1+4/5=8+4/5=44/5
\(\frac{4}{3}+\frac{7}{6}+\frac{11}{10}+...+\frac{46}{45}\)
\(=1+\frac{1}{3}+1+\frac{1}{6}+1+\frac{1}{10}+...+1+\frac{1}{45}\)
\(=\left(1+1+1+...+1\right)+\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)(8 chữ số 1)
\(=8+\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)
Đặt A = \(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\)
=> \(\frac{1}{2}\)A = \(\frac{1}{2}\times\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{45}\right)\)
= \(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{90}\)
\(=\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{9.10}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{9}-\frac{1}{10}\)
\(=\frac{1}{2}-\frac{1}{10}=\frac{2}{5}\)
Vậy A = \(\frac{2}{5}:\frac{1}{2}=\frac{4}{5}\)
Do đó biểu thức trên là 8 + \(\frac{4}{5}\) = \(\frac{44}{5}\)
Đáp số: \(\frac{44}{5}\)
a) $\frac{1}{2} - \frac{1}{4} = \frac{2}{4} - \frac{1}{4} = \frac{1}{4}$
b) $\frac{2}{3} - \frac{4}{{15}} = \frac{{10}}{{15}} - \frac{4}{{15}} = \frac{6}{{15}} = \frac{2}{5}$
c) $\frac{3}{5} - \frac{{10}}{{25}} = \frac{{15}}{{25}} - \frac{{10}}{{25}} = \frac{5}{{25}} = \frac{1}{5}$
C=\(\frac{1}{3}+\frac{1}{10}+\frac{1}{15}+\frac{1}{20}+\frac{1}{25}+\frac{1}{30}+\frac{1}{35}+\frac{1}{40}\)
=\(\frac{1}{3}+\frac{1}{15}+\frac{1}{10}+\frac{1}{20}+\frac{1}{30}+\frac{1}{40}+\frac{1}{25}+\frac{1}{35}\)
=\(\frac{5}{15}+\frac{1}{15}+\frac{4}{40}+\frac{2}{40}+\frac{1}{40}+\frac{1}{30}+\frac{1}{25}+\frac{1}{35}\)
=\(\frac{6}{15}+\frac{7}{40}+\frac{107}{1050}\)