Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Ta có : \(\frac{3\cdot5\cdot7\cdot11\cdot13\cdot37-10101}{1212120+40404}\)
\(=\frac{(3\cdot7\cdot11\cdot13\cdot37)\cdot5-10101\cdot1}{120\cdot10101+4\cdot10101}\)
\(=\frac{10101\cdot(5-1)}{10101\cdot(120+4)}=\frac{4}{124}=\frac{1}{31}\)
~Chúc bạn học tốt~
Ta có:
\(A=\frac{3.5.7.11.13.37-10101}{1212120+40404}\)
\(=\frac{\left(3.7.11.13.37\right).5-10101.1}{120.10101+4.10101}\)
\(=\frac{10101.\left(5-1\right)}{10101.\left(120+4\right)}\)
\(=\frac{4}{124}=\frac{1}{31}\)
Sai rồi:
A = 5.11.(3.7.13.37) - 10101/(10101.120 + 10101.4)
= (5.11.10101 - 10101)/(10101.120+10101.4)
= 10101(5.11-1)/10101(120+4)
= 27/62.
\(=\frac{3\text{x}5x7x11\text{x}13\text{x}37-3\text{x}7\text{x}13\text{x}37}{2^3x3^2x5\text{x}7\text{x}13\text{x}37+2^2x3\text{x}7\text{x}13\text{x}37}=\frac{3\text{x}7\text{x}13\text{x}37\left(5\text{x}11-1\right)}{2^2x3\text{x}7\text{x}13\text{x}37\left(2\text{x}3\text{x}5+1\right)}\)
\(=\frac{54}{2^2x31}=\frac{54}{124}=\frac{27}{62}\) thử lại kết quả rồi nhé
1-3-5+7+9-11-13+15+...+2017-2019-2021+2023=
=(1-3-5+7)+(9-11-13+15)+...+(2017-2019-2021+2023)=
=0+0+.....+0=0
\(=\dfrac{10}{17}-\dfrac{5}{3}+7-\dfrac{8}{13}+\dfrac{11}{25}\)
\(=\dfrac{30}{51}-\dfrac{85}{51}+\dfrac{175}{25}+\dfrac{11}{25}-\dfrac{8}{13}\)
\(=\dfrac{-55}{51}+\dfrac{186}{25}-\dfrac{8}{13}\)
\(=\dfrac{-55\cdot325+186\cdot663-8\cdot1275}{16575}=\dfrac{95243}{16575}\)
.5.7.11.13.37-10101/1212120+40404
=3.5.7.11.13.37-10101/1212120.1/10+40404 (vì 1/1212120=1/121212.1/10)
= 3.37.7.11.13.5-101010/121212.1/100+40404
=111.1001.5-5/6.1/100+40404
=151515.5-250/3
=595959-250/3
=1787876/3
A= [5.11.10101 -10101]/[10101.120+10101.4] = 10101.[5.11-1] / 101.[120+4] = 54/124=27/62
đúng nhé
5.7.11.13.37-10101/1212120+40404
=3.5.7.11.13.37-10101/1212120.1/10+40404 (vì 1/1212120=1/121212.1/10)
= 3.37.7.11.13.5-101010/121212.1/100+40404
=111.1001.5-5/6.1/100+40404
=151515.5-250/3
=595959-250/3
=1787876/3