S= 3/5.7+3/7.9+...+3/59.61
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Giải:
S=3/5.7+3/7.9+...+3/59.61
S=3/2.(2/5.7+2/5.7+...+2/59.61)
S=3/2.(1/5-1/7+1/7-1/9+...+1/59-1/61)
S=3/2.(1/5-1/61)
S=3/2.56/305
S=84/305
Chúc bạn học tốt!
S= 3/5.7 + 3/7.9 +...........................+3/59.61
=3/2.(1/5 - 1/7 +1/7 -1/9 +....................+ 1/59-1/61)
=3/2.(1/5-1/61)
mình chỉ làm được tới đó
= 3(1/5.7+1/7.9+...+1/59.61)
= 3/2(2/5.7+2/7.9+...+2/59.61)
= 3/2(1-1/5+1/5-1/7+1/7-1/9+...+1/59-1/61)
= 3/2(1-1/61)=3/2.60/61=90/61
Chẳng biết mk làm đúng ko nữa!
\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}=\frac{3}{2}\cdot\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{84}{305}\)
gọi biểu thức trên là A. ta có:
3A = 1/5.7+1/7.9+......+ 1/59.61
3A = 1/5-1/7+1/7-1/9+....+1/59-1/61
3A = 1/5 - 1/61
3A = 56/305
A = 56/305 : 3 = 56/915
\(M=\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{59\cdot61}\)
\(M=\frac{3}{2}\left[\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{59\cdot61}\right]\)
\(M=\frac{3}{2}\left[\frac{1}{5}-\frac{1}{7}+...+\frac{1}{59}-\frac{1}{61}\right]\)
\(M=\frac{3}{2}\left[\frac{1}{5}-\frac{1}{61}\right]\)
\(M=\frac{3}{2}\cdot\frac{56}{305}=\frac{84}{305}\)
.Cho P=3/5.7+3/7.9+...+3/59.61=3/2.(1/5-1/61)=3/2.56/305=84/305
Đặt \(A=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(\Rightarrow A=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)
\(S=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{3}{2}\left(\frac{61-5}{305}\right)=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)