\(\dfrac{1}{1.4}+\dfrac{1}{4.7}+...+\dfrac{1}{34.37}+\dfrac{1}{37.40}\)

\(=1-\left(\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\right)\)

\(=1-\dfrac{1}{40}\)

\(=\dfrac{39}{40}\)

Vậy giá trị cần tìm là: \(\dfrac{39}{40}\)

\(3xA=\dfrac{4-1}{1x4}+\dfrac{7-4}{4x7}+...+\dfrac{37-34}{34x37}+\dfrac{40-37}{37x40}=\)

\(=1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{17}{ }+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}=\)

\(=1-\dfrac{1}{40}=\dfrac{39}{40}\Rightarrow A=\dfrac{39}{40}:3=\dfrac{13}{40}\)

\(=\dfrac{1}{3}\left(\dfrac{3}{1.4}+\dfrac{3}{4.7}+...+\dfrac{3}{34.37}+\dfrac{3}{37.40}\right)\)

\(=\dfrac{1}{3}\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{34}-\dfrac{1}{37}+\dfrac{1}{37}-\dfrac{1}{40}\right)\)

\(=\dfrac{1}{3}.\dfrac{39}{40}\)

\(=\dfrac{13}{40}\)

\(A=\dfrac{1}{1\times4}+\dfrac{1}{4\times7}+...+\dfrac{1}{37\times40}\\ =\dfrac{1}{3}\times\left(\dfrac{3}{1\times4}+\dfrac{3}{4\times7}+...+\dfrac{3}{37\times40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+...+\dfrac{1}{37}-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\left(1-\dfrac{1}{40}\right)\\ =\dfrac{1}{3}\times\dfrac{39}{40}\\ =\dfrac{13}{40}\)

=1-1/4+1/4-1/7+1/7-...+1/37-1/40

=1-1/40=39/40

Mình cần gấp ạ. Mốt mik thi rồi

Mình đang gấp lắm ạ

\(\frac{2}{1\times4}+\frac{2}{4\times7}+\frac{2}{7\times10}+...+\frac{2}{37\times40}\)

\(=\frac{2}{3}\times\left(\frac{3}{1\times4}+\frac{3}{4\times7}+\frac{3}{7\times10}+...+\frac{3}{37\times40}\right)\)

\(=\frac{2}{3}\times\left(\frac{4-1}{1\times4}+\frac{7-4}{4\times7}+\frac{10-7}{7\times10}+...+\frac{40-37}{37\times40}\right)\)

\(=\frac{2}{3}\times\left(1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{37}-\frac{1}{40}\right)\)

\(=\frac{2}{3}\times\left(1-\frac{1}{40}\right)=\frac{13}{20}\)

\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)

\(\dfrac{8}{7}A=\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\)

\(\dfrac{8}{7}A=1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\)

\(\dfrac{8}{7}A=1-\dfrac{1}{89}=\dfrac{88}{89}\Rightarrow A=\dfrac{88}{89}:\dfrac{8}{7}=\dfrac{77}{89}\)

\(B=\dfrac{5^2}{1.4}+\dfrac{3^2}{4.7}+\dfrac{3^2}{7.10}+...+\dfrac{3^2}{37.40}\)

\(B=\dfrac{25}{1.4}+\dfrac{9}{4.7}+\dfrac{9}{7.10}+...+\dfrac{9}{37.40}\)

\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{3}{4.7}+\dfrac{3}{7.10}+...+\dfrac{3}{37.40}\)

\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+...+\dfrac{1}{37}-\dfrac{1}{40}\)

\(\dfrac{1}{3}B=\dfrac{25}{12}+\dfrac{1}{4}-\dfrac{1}{40}=\dfrac{277}{120}\Rightarrow B=\dfrac{277}{120}:\dfrac{1}{3}=\dfrac{277}{40}\)

\(A=\dfrac{7}{1.9}+\dfrac{7}{9.17}+\dfrac{7}{17.25}+...+\dfrac{7}{81.89}\)

\(=7\left(\dfrac{8}{1.9}+\dfrac{8}{9.17}+\dfrac{8}{17.25}+...+\dfrac{8}{81.89}\right)\)

\(=7\left(1-\dfrac{1}{9}+\dfrac{1}{9}-\dfrac{1}{17}+\dfrac{1}{17}-\dfrac{1}{25}+\dfrac{1}{25}+...+\dfrac{1}{81}-\dfrac{1}{89}\right)\)

\(=7.\left(1-\dfrac{1}{89}\right)=7.\dfrac{88}{89}=\dfrac{616}{89}\)

\(\dfrac{m}{n}=\dfrac{7}{1\cdot4}+\dfrac{7}{4\cdot7}+...+\dfrac{7}{37\cdot40}\)

\(=\dfrac{7}{3}\left(\dfrac{3}{1\cdot4}+\dfrac{3}{4\cdot7}+...+\dfrac{3}{37\cdot40}\right)\)

\(=\dfrac{7}{3}\left(1-\dfrac{1}{40}\right)\)

\(=\dfrac{7}{3}\cdot\dfrac{39}{40}=\dfrac{91}{40}\)

\(\Leftrightarrow\left(m,n\right)=\left(91;40\right)\)

Suy ra: S=91+40=131

xet 1/4x7 +1/7x10+...+1/37x40

dat bieu thuc tren la A ta co:

A=1/4x7+1/7x10+...+1/37x40

3A=3/4X7+3/7X10+...+3/37X40

3A=1/4-1/71/7-1/10+...+1/37-1/40

3A=1/4-1/40

3A=10/40-1/40=9/40

A=9/40:3=9/120=3/40

=> (1/4x7+1/7x10+...+1/37x40)-x=4/5

3/40-x=4/5

   x=3/40-4/5=-29/40

Ta có    1/1x4+1/4x7+...+1/2002x2005

  <=>   =1/3.3(1/1x4+1/4x7+...+1/2002x2005)

            =1/3(3/1x4+3/4x7+...+3/2002x2005)

            =1/3(1-1/4+1/4-1/7+...+1/2002-1/2005)

            =1/3(1-1/2005)

            =1/3.2004/2005

            =1.2004/3.2005

            =668/2005

\(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+...+\(\frac{1}{2002.2005}\)=3(\(\frac{1}{1.4}\)+ \(\frac{1}{4.7}\)+...+ \(\frac{1}{2002.2005}\)):3=(\(\frac{3}{1.4}\)+ \(\frac{3}{4.7}\)+...+ \(\frac{3}{2002.2005}\)):3= (1-\(\frac{1}{4}\)+ \(\frac{1}{4}\)- \(\frac{1}{7}+...+\frac{1}{2002}-\frac{1}{2005}\)):3=(1-\(\frac{1}{2005}\)) : 3 = \(\frac{668}{2005}\)

= \(\frac{1}{3}\left(\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-.....+\frac{1}{x}-\frac{1}{x+1}\right)\) ) 

= \(\frac{1}{3}\left(1-\frac{1}{x+1}\right)\)

=\(\frac{1}{3}x\left(\frac{x+1}{x+1}-\frac{1}{x+1}\right)\)

= \(\frac{1}{3}x\frac{x}{x+1}\)

dè bài la tim x

Tong cua phep. Tinh la 2009/6030