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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
\(=4\left(\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{53.55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{5}+\frac{1}{7}-\frac{1}{7}+...+\frac{1}{53}-\frac{1}{55}\right)\)
\(=4\left(\frac{1}{5}-\frac{1}{55}\right)\)
\(=4.\frac{2}{11}\)
\(=\frac{8}{11}\)
Đặt A=\(\dfrac{2}{3.5}.\dfrac{2}{7.9}.....\dfrac{2}{99.101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
A=\(\dfrac{1}{3}-\dfrac{1}{101}=\dfrac{98}{303}\)
Ta có: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
N=3/2.(1/5.7+1/7.9+1/9.11+....+1/197.199)
N=3/2.(1/5-1/7+1/7-1/9+1/9-1/11+....+1/197-1/199)
N=3/2.(1/5-1/199)
N=3/2.194/995
N=291/995
Đặt 2/5 ra ngoài rồi tách từng cặp phân số ra sau đó bn tự làm nhé!
A=\(\frac{5}{5.7}+\frac{5}{7.9}+.........+\frac{5}{59.61}\)
=\(\frac{5}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+.........+\frac{2}{59.61}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...........+\frac{1}{59}-\frac{1}{61}\right)\)
=\(\frac{5}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)
=\(\frac{5}{2}.\frac{56}{305}\)
=\(\frac{28}{61}\)
= 5-7+7-9+9-11+...+25-27
= 5-27
= -22
5.7+7.9+9.11+...+25.27
=5.(5+2)+7.(7+2)+9.(9+2)+...+25.(25+2)
=52+2.5+72+2.7+92+2.9+...+252+2.25
=(52+72+92+...+252)+(2.5+2.7+2.9+...+2.25)
=2915+2.(5+7+9+...+25)
=2915+2.165
=2915+330
=3245