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

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

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

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

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

=> 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= \(\frac{84}{305}\)

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

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

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

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

    \(=\frac{3}{2}.\frac{56}{305}\)

    \(=\frac{84}{305}\)

biểu thức trên =\(\frac{1}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+.....+\frac{1}{56}-\frac{1}{61}\right)=\frac{1}{2}\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{1}{2}x\frac{61}{305}=\frac{1}{10}=0,1.\)

vậy biểu thức trên =0,1

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

\(\frac{2}{3}A=\frac{2.3}{3.5.7}+\frac{2.3}{3.7.9}+...+\frac{2.3}{3.59.61}\)

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

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

\(\frac{2}{3}A=\frac{1}{5}-\frac{1}{61}\)

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

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

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

dễ ợt

A= 1/5-1/61= 56/305

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

\(A=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{3}{59.61}\)

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

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

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

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

\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{84}{305}\)

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

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

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

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

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

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

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

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

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

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 =\(\frac{1105}{144}\)

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

Ta có 1/2B=2/5.7+2/7.9+...+2/59.61

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

1/2B=1/5-1/61

1/2B=56/305

B=56/305:1/2

B=112/305