Tính hợp lí
a) 3 5.7 + 3 7.9 + ... + 3 59.61
b) 5 22 + 3 13 − 1 2 4 13 − 2 11 + 3 2
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NT
2
14 tháng 7 2021
a) Ta có: \(\dfrac{3}{5\cdot7}+\dfrac{3}{7\cdot9}+...+\dfrac{3}{59\cdot61}\)
\(=\dfrac{3}{2}\left(\dfrac{2}{5\cdot7}+\dfrac{2}{7\cdot9}+...+\dfrac{2}{59\cdot61}\right)\)
\(=\dfrac{3}{2}\left(\dfrac{1}{5}-\dfrac{1}{61}\right)\)
\(=\dfrac{3}{2}\cdot\dfrac{56}{305}=\dfrac{84}{305}\)
2 tháng 4 2019
a.\(4\frac{3}{4}+\left(-0,37\right)+\frac{1}{8}+\left(-1,28\right)+\left(2,5\right)+3\frac{1}{12}\)
\(=\)\(\left(4\frac{3}{4}+\frac{1}{8}+3\frac{1}{12}\right)-\left(0,37+1,28+2,5\right)\)
\(=7\frac{23}{24}-4,15\)
\(=7\frac{23}{24}-4\frac{3}{20}\)
\(=3\frac{97}{120}\)
b.\(\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(=\frac{3}{2}\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...\frac{1}{59}-\frac{1}{61}\right)\)
\(=\frac{3}{2}\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\frac{56}{305}\)
\(=\frac{84}{305}\)
c.\(\frac{\frac{5}{22}+\frac{3}{13}-\frac{1}{2}}{\frac{4}{13}-\frac{2}{11}+\frac{3}{2}}\)
\(=\frac{\left(\frac{5}{22}+\frac{3}{13}-\frac{1}{2}\right).\left(2.11.13\right)}{\left(\frac{4}{13}-\frac{2}{11}+\frac{3}{2}\right).\left(2.11.13\right)}\)
\(=\frac{65+66-143}{88-52+429}\)
\(=\frac{-12}{465}=\frac{-4}{155}\)
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
\(2\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=2\left(\frac{1}{5}-\frac{1}{61}\right)=2\left(\frac{61-5}{305}\right)=2.\frac{56}{305}=\frac{112}{305}\)
11 tháng 5 2015
Đặt \(A=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(\Rightarrow A=\frac{3}{2}.\left(\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{59.61}\right)=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)=\frac{3}{2}.\frac{56}{305}=\frac{84}{305}\)
T
2
T
11 tháng 5 2018
M=3.(\(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+\frac{1}{9}-....+\frac{1}{59}-\frac{1}{60}\)\(\frac{1}{61}\))
M= 3.(\(\frac{1}{5}-\frac{1}{61}\))
M=\(\frac{168}{305}\)
11 tháng 5 2018
\(M=\frac{3}{5.7}+\frac{3}{7.9}+...+\frac{3}{59.61}\)
\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+...+\frac{1}{59}-\frac{1}{61}\right)\)
\(M=\frac{3}{2}.\left(\frac{1}{5}-\frac{1}{61}\right)\)
\(M=\frac{84}{305}\)
a)
3 5.7 + 3 7.9 + ... + 3 59.61 = 3 2 . 2 5.7 + 2 7.9 + ... + 2 59.61 = 3 2 . 1 5 − 1 7 + 1 7 − 1 9 + ... + 1 59 − 1 61 = 3 2 . 1 5 − 1 61 = 3 2 . 56 305 = 844 305
b)
5 22 + 3 13 − 1 2 4 13 − 2 11 + 3 2 = 5 22 + 3 13 − 1 2 . 2.11.13 4 13 − 2 11 + 3 2 . 2.11.13 = 65 + 66 − 143 88 − 52 + 429 = − 12 465 = − 4 155