N = \(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{197.199}\)
(nhớ có lời giải nhé)
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\(N=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{197}-\frac{1}{199}\right)\)
\(N=\frac{3}{2}\cdot\left(\frac{1}{5}-\frac{1}{199}\right)\)
\(N=\frac{3}{2}\cdot\frac{194}{995}\)
\(N=\frac{291}{995}\)
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)=3/2.194/995=291/995
\(\frac{2}{3}N=\frac{1}{5.7}+\frac{1}{7.9}+\frac{1}{9.11}+...+\frac{1}{197.199}\)
\(=\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-\frac{1}{11}+...+\frac{1}{197}-\frac{1}{199}\)
\(=\frac{1}{5}-\frac{1}{199}\)
\(=\frac{194}{995}\)
=> \(N=\frac{194}{995}:\frac{2}{3}=\frac{291}{995}\)
BÀi này dễ thôi bạn ạ
N=3(1/5.7+1/7.9+.........+1/197.199)
N=3/2( 1/5-1/7+1/7-1/9+1/9-..........+1/197-1/199)
N=3/2(1/5-1/199)
N=3/2.194/995
N=291/995
k đúng cho mình nhé
N=3.1/2.(1/5-1/7+1/7-1/9+1/9-1/11+...+1/197-1/199)
N=3.1/2.(1/5-1/99)
N=3.1/2.94/495
N=3.47/495
N=47/165
N = ( 3/5 - 3/7 + 3/7 - 3/9 + 3/9 - 3/11 + ... + 3/197 - 3/199 ) : 2
N = ( 3/5 - 3/199 ) : 2
N = 582/995 : 2
N = 291/ 995
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
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
N=3/5.7+3/7.9+3/9.11+..................+3/197.199
N=1/5-1/7+1/7-1/9+...........+1/197.199
N=1/5-1/199
=194/995
a)\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{2013}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{2015}\right)\)
\(=\frac{3}{2}\cdot\frac{402}{2015}\)
\(=\frac{603}{2015}\)
b)\(=\frac{4}{5}\left(\frac{1}{3}-\frac{1}{8}+\frac{1}{8}-\frac{1}{13}+...+\frac{1}{93}-\frac{1}{98}\right)\)
\(=\frac{4}{5}\left(\frac{1}{3}-\frac{1}{98}\right)\)
\(=\frac{4}{5}\cdot\frac{95}{294}\)
\(=\frac{38}{147}\)
a) Gọi tổng trên là A
A = \(\frac{3}{5.7}+\frac{3}{7.9}+\frac{3}{9.11}+...+\frac{3}{2013.2015}\)
A == \(\frac{3}{5}-\frac{3}{7}+\frac{3}{7}-\frac{3}{9}+\frac{3}{9}-\frac{3}{11}+...+\frac{3}{2013}-\frac{3}{2015}\)
Vì một số trừ cho a rồi cộng cho a sẽ bằng chính số đó nên:
A = \(\frac{3}{5}-\frac{3}{2015}\)
A = \(\frac{1209}{2015}-\frac{3}{2015}\)
A = \(\frac{1206}{2015}\)
b) Gọi tổng trên là B
B = \(\frac{4}{3.8}+\frac{4}{8.13}+\frac{4}{13.15}+...+\frac{4}{93.98}\)
B = \(\frac{4}{3}-\frac{4}{8}+\frac{4}{8}-\frac{4}{13}+\frac{4}{13}-\frac{4}{15}+...+\frac{4}{93}-\frac{4}{98}\)
Vì một số trừ cho a rồi cộng cho a sẽ bằng chính số đó nên:
B = \(\frac{4}{3}-\frac{4}{98}\)
B = \(\frac{686}{294}-\frac{12}{294}\)
B = \(\frac{674}{294}=\frac{337}{147}\)
2\3x-780\11:[13\2.(1\3.5+1\5.7+1\7.9+1\9.11]=-5
2\3x-780\11:[13\2.(1\3-1\5+1\5-1\7+....+1\9-1\11)]=-5
2\3x-780\11:[13\2.(1\3-1\11)]=-5
2\3x-780\11:[13\2.8\33]=-5
2\3x-780\11:52\33=-5
2\3x-525\13=-5
2\3x=-5+525\13
2\3x=460\13
x=460\13:2\3
x=690\13
N=\(3.\left(\frac{1}{5.7}+\frac{1}{7.9}+...+\frac{1}{197.199}\right)\)
N=\(\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{197.199}\right)\)
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=\(\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{999}\right)\)
N=\(\frac{3}{2}.\frac{194}{995}\)
N=\(\frac{291}{995}\)
= 3/5-3/7+3/7-3/9+3/9-3/11+......+3/197-3/199
= 3/5-3/199
= 582/995