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CQ
0
NT
2
14 tháng 7 2021
a) Ta có: \(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)
LT
0
3 tháng 4 2019
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
27 tháng 2 2015
a,Gọi tổng trên là A.
Xét \(\frac{4}{5}-\frac{4}{7}=\frac{8}{35};...;\frac{4}{59}-\frac{4}{61}=\frac{8}{3599}\)=>\(A=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{7}+\frac{4}{7}-\frac{4}{9}+...+\frac{4}{59}-\frac{4}{61}\right)\)\(=\frac{1}{2}.\left(\frac{4}{5}-\frac{4}{61}\right)=\frac{1}{2}.\frac{224}{305}=\frac{112}{305}\)
b,Gọi tổng trên là B
Theo đề bài ta có:\(B=\frac{24.47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}\)=\(\frac{\left(23+1\right).47-23}{24+47.23}.\frac{3+\frac{3}{7}-\frac{3}{11}+\frac{3}{1001}-\frac{3}{13}}{\frac{9}{1001}-\frac{9}{13}+\frac{9}{7}-\frac{9}{11}+9}=\frac{47.23+24}{24+47.23}.\frac{3.\left(1+\frac{1}{7}-\frac{1}{11}+\frac{1}{1001}-\frac{1}{13}\right)}{3.\left(3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}\right)}\)\(=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3+\frac{3}{1001}-\frac{3}{13}+\frac{3}{7}-\frac{3}{11}}=\frac{1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}}{3.\left(1+\frac{1}{1001}-\frac{1}{13}+\frac{1}{7}-\frac{1}{11}\right)}=\frac{1}{3}\)
LP
29 tháng 2 2016
\(2\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{61}\right)=2\left(\frac{61-5}{305}\right)=2.\frac{56}{305}=\frac{112}{305}\)
PK
1
13 tháng 6 2017
\(A=\dfrac{6^{12}.3^3.5-7.9^7.2^{13}}{2.4^7.5-2^{14}.3^2}\)
\(=\dfrac{2^{12}.3^{12}.3^3.5-7.3^{14}.2^{12}}{2.2^{14}.5-2^{14}.3^2}\)
\(=\dfrac{2^{12}.3^{15}.5-7.3^{14}.2^{13}}{2.3^{14}.5-2^{14}.3^2}\)
\(=\dfrac{2^{12}.3^{14}.\left(15-14\right)}{2^{14}.\left(2.5-3\right)}\)
\(=\dfrac{3^{14}.1}{2^2}\)
\(=\dfrac{314}{4}\)
TH
1
=-3/7.(4/13+9/13)+3/7
=-3/7.13/13+3/7
=-3/7.1+3/7
=-3/7+3/7
=0