A=5.7+7.9+...+31.33
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\(\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}+...+\frac{2}{31.33}+\frac{2}{33.35}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{31}-\frac{1}{33}+\frac{1}{33}-\frac{1}{35}\)
\(=\frac{1}{5}-\frac{1}{35}\)
\(=\frac{6}{35}\)
Chúc bạn học giỏi nha!!!
K cho mik vs nhé danggiahuy
đặt A=2/5.7+2/7.9+2/9.11+.....+2/31.33+2/33.35
A=1/5-1/7+1/7-1/9+1/9-1/11+.....+1/31-1/33+1/33-1/35
A=1/5-1/35
A=6/35
\(\frac{5}{3.5}\)+ \(\frac{5}{5.7}\)+ ..... + \(\frac{5}{31.33}\)
= \(\frac{5}{2}\). ( \(\frac{1}{3}\)_ \(\frac{1}{5}\)+ \(\frac{1}{5}\)_ \(\frac{1}{7}\)+.....+ \(\frac{1}{31}\)_ \(\frac{1}{33}\))
= \(\frac{5}{2}\). ( \(\frac{1}{3}\)_ \(\frac{1}{33}\))
= \(\frac{5}{2}\). \(\frac{30}{99}\)
= \(\frac{50}{66}\)
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
\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{201.203}\)
\(=\frac{1}{2}.2.\left(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{201.203}\right)\)
\(=\frac{1}{2}.2.3.\left(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{201.203}\right)\)
\(=\left(\frac{1}{2}.3\right).2.\left(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{201.203}\right)\)
\(=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{201.203}\right)\)
Vì muốn chuyển 3/5.7 = 1/5 - 1 /7 thì tử số phải bằng hiệu của mẫu số nên 3/5.7= 3/5.7 chia 2/5.7 = 3/2 . 2/5.7 các phân số khác cũng tương tự như thế
nên ta có 3/5.7 +3/7.9 +...3/201.203 = 3/2. (2/5.7+2/7.9+...+2/201.203)
\(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{97.99}\)
\(=\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{97}-\frac{1}{99}\)
\(=\frac{1}{3}+\left(\frac{1}{5}-\frac{1}{5}\right)+\left(\frac{1}{7}-\frac{1}{7}\right)+...+\left(\frac{1}{97}-\frac{1}{97}\right)-\frac{1}{99}\)
\(=\frac{1}{3}-\frac{1}{99}=\frac{32}{99}\)
~ Hok tốt ~
\(\)
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: \(P=\dfrac{2}{3\cdot5}+\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+\dfrac{2}{9\cdot11}+\dfrac{2}{11\cdot13}+\dfrac{2}{13\cdot15}\)
\(=\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{13}-\dfrac{1}{15}\)
\(=\dfrac{1}{3}-\dfrac{1}{15}\)
\(=\dfrac{4}{15}\)
A=(31.33-5):2+1x7
còn gì thì tự tính nốt nha