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\(T=\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+\dfrac{3}{9\cdot11}+...+\dfrac{3}{59\cdot61}\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+...+\dfrac{2}{59\cdot61}\right)\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+...+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}\cdot\left(\dfrac{1}{5}-\dfrac{1}{61}\right)=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)
\(\dfrac{3}{5.7}+\dfrac{3}{7.9}+\dfrac{3}{9.11}+...+\dfrac{3}{59.61}\)
\(=3.\left(\dfrac{1}{5.7}+\dfrac{1}{7.9}+\dfrac{1}{9.11}+...+\dfrac{1}{59.61}\right)\)
\(=3.\dfrac{1}{2}.\left(\dfrac{1}{5}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{9}+....+\dfrac{1}{59}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}.\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}.\dfrac{56}{305}\)
\(=\dfrac{84}{305}\)
Đặt 2/5 ra ngoài rồi tách từng cặp phân số ra sau đó bn tự làm nhé!
A=\(\frac{5}{5.7}+\frac{5}{7.9}+.........+\frac{5}{59.61}\)
=\(\frac{5}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+.........+\frac{2}{59.61}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...........+\frac{1}{59}-\frac{1}{61}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)
=\(\frac{5}{2}.\frac{56}{305}\)
=\(\frac{28}{61}\)
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!
= 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!
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
\(M=\frac{3}{5\cdot7}+\frac{3}{7\cdot9}+...+\frac{3}{59\cdot61}\)
\(M=\frac{3}{2}\left[\frac{2}{5\cdot7}+\frac{2}{7\cdot9}+...+\frac{2}{59\cdot61}\right]\)
\(M=\frac{3}{2}\left[\frac{1}{5}-\frac{1}{7}+...+\frac{1}{59}-\frac{1}{61}\right]\)
\(M=\frac{3}{2}\left[\frac{1}{5}-\frac{1}{61}\right]\)
\(M=\frac{3}{2}\cdot\frac{56}{305}=\frac{84}{305}\)
a.
\(M=1.\left[\frac{1}{3}-\frac{1}{5}+.....\frac{1}{97}-\frac{1}{99}\right]\)
\(M=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
b.
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{197}-\frac{1}{199}\right]\)
\(N=\frac{3}{2}.\left[\frac{1}{5}-\frac{1}{199}\right]=\frac{291}{995}\)
mk đầu tiên nha bạn
84/305 nha bạn
mk nha có cần giải ra ko