Đặt A=như đã cho.

=>1/2A=2/5*7+2/7*9+2/9*11+...+2/59*61.

=>1/2A=1/5-1/7+1/7-1/9+1/9-1/11+...+1/59-1/61.

=>1/2A=1/5-1/61=56/305.

=>A=56/305*2=112/305.

k nha đúng đó.Có j kb nha.

pn kia nhầm zùi

Bài này dễ thôi mà bạn

kho qua tra loi ho nha

= 2/2 . ( 4 / 5.7 +4 / 7.9 +...+ 4 / 59.61 ) 

= 4/2 . ( 2 / 5.7 +2 / 7.9 +...+ 2 / 59.61 )

= 2 . ( 7-5 / 5.7 + 9-7 / 7.9 +...+ 61-59 / 59.61 )

= 2 . ( 1/5 - 1/7 + 1/7 - 1/9 +...+ 1/59 - 1/61 )

= 2 . ( 1/5 - 1/61 )

= 2 . ( 61/305 - 5/305 ) 

= 2 . 56/305

= 112/305 

ảnh anime đẹp thế là anime gì vậy bạn

JUST A WORD UP! PLEASE !

Ta có:\(\frac{4}{5.7}+\frac{4}{7.9}+.....+\frac{4}{59.61}\)

\(\Rightarrow2.\left(\frac{2}{5.7}+\frac{2}{7.9}+......+\frac{2}{59.61}\right)\)

\(\Rightarrow2.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{59}-\frac{1}{61}\right)\)

\(\Rightarrow2.\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(\Rightarrow\frac{112}{305}\)

\(\frac{4}{5.7}+\frac{4}{7.9}+...+\frac{4}{59.61}\)

\(=\frac{4.2}{5.7.2}+\frac{4.2}{7.9.2}+...+\frac{4.2}{59.61.2}\)

\(=\frac{4}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

\(=\frac{4}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}_{ }\right)\)

\(=\frac{4}{2}.\left(\frac{1}{5}-\frac{1}{60}\right)\)

\(=\frac{4}{2}.\frac{11}{60}\)

\(=\frac{11}{30}\)

\(\frac{\text{4}}{5.7}\)+ \(\frac{4}{7.9}\)+...+ \(\frac{4}{59.60}\)

=\(\frac{4}{2}\).( \(\frac{2}{5.7}\)+\(\frac{2}{7.9}\)+...+\(\frac{2}{59.60}\))

=\(\frac{4}{2}\).(\(\frac{1}{5}\)-\(\frac{1}{7}\)+\(\frac{1}{7}\)-\(\frac{1}{9}\)+...+\(\frac{1}{59}\)-\(\frac{1}{60}\))

=\(\frac{4}{2}\).(\(\frac{1}{5}\)-\(\frac{1}{60}\))

=\(\frac{2}{5}\)-\(\frac{1}{30}\)

=\(\frac{12}{30}\)-\(\frac{1}{30}\)

=\(\frac{11}{30}\)

\(=\frac{1}{5}-\frac{1}{61}=\frac{56}{305}\)

\(S=\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\)

\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)

\(S=\frac{1}{5}-\frac{1}{61}\)

\(S=\frac{61}{305}-\frac{5}{305}\)

\(S=\frac{56}{305}\)

\(S=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\)

\(S=\frac{1}{5}-\frac{1}{61}\)

\(S=\frac{56}{305}\)

a,Gọi tổng trên là A.

Xét \(\frac{4}{5}-\frac{4}{7}=\frac{8}{35};...;\frac{4}{59}-\frac{4}{61}=\frac{8}{3599}\)=>\(A=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{59}-\frac{4}{61}\right)\)\(=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{61}\right)=\frac{1}{2}.\frac{224}{305}=\frac{112}{305}\)

b,Gọi tổng trên là B

Theo đề bài ta có:\(B=\frac{24.47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)=\(\frac{\left(23+1\right).47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}=\frac{47.23+24}{24+47.23}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{3.\left(3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}\right)}\)\(=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}}=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3.\left(1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{1}{3}\)

\(2\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(=2\left(\frac{1}{5}-\frac{1}{61}\right)=2\left(\frac{61-5}{305}\right)=2.\frac{56}{305}=\frac{112}{305}\)

Đặt A=B*C

B=\(\frac{24\cdot47-23}{24+47-23}=\frac{1128-23}{71-23}=\frac{1105}{48}\)

C=\(\frac{3\cdot\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{9\cdot\left(\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}+1\right)}=\frac{1}{3}\)

Suy ra A =\(\frac{1105}{144}\)

\(\Rightarrow A=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+....+\frac{2}{59.61}\right)\)

\(\Rightarrow A=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+......+\frac{1}{59}-\frac{1}{61}\right)\)

\(\Rightarrow A=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(\Rightarrow A=\frac{3}{2}.\frac{56}{305}\)

\(\Rightarrow A=\frac{84}{305}\)

=> A= \(\frac{3}{2}\) .( \(\frac{1}{5}\) - \(\frac{1}{7}\) + \(\frac{1}{7}\) - \(\frac{1}{9}\) +...+ \(\frac{1}{59}\) - \(\frac{1}{61}\))

=> A=\(\frac{3}{2}\) . (\(\frac{1}{5}\) - \(\frac{1}{61}\) ) => A= \(\frac{3}{2}\). \(\frac{56}{305}\) = \(\frac{84}{305}\) Vậy A= \(\frac{84}{305}\)

\(A=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)

    \(=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)

    \(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{53}-\frac{1}{61}\right)\)

    \(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)

    \(=\frac{3}{2}.\frac{56}{305}\)

    \(=\frac{84}{305}\)

M=3.(\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-....+\frac{1}{59}-\frac{1}{60}\)\(\frac{1}{61}\))

M= 3.(\(\frac{1}{5}-\frac{1}{61}\))

M=\(\frac{168}{305}\)

\(M=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)

\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)

\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)

\(M=\frac{84}{305}\)