3.5.7.11.13.37-10101 / 1212120+40404
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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
a: \(\dfrac{31995-81}{42660-108}=\dfrac{31914}{42552}=\dfrac{3}{4}\)
b: \(\dfrac{3\cdot5\cdot7\cdot11\cdot13\cdot37-10101}{1212120+40404}=\dfrac{545454}{1252524}=\dfrac{27}{62}\)
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.
\(\Rightarrow\)A = \(\dfrac{\left(3.7.13.37\right).\left(5.11\right)-10101.1}{40404.30+40404.1}\)
\(\Rightarrow\)A = \(\dfrac{10101.55-10101.1}{40404.\left(30+1\right)}\)
\(\Rightarrow\)A =\(\dfrac{10101.\left(55-1\right)}{10101.\left(4.30\right)}\)
\(\Rightarrow\)A= \(\dfrac{10101.54}{10101.124}\)
chia cả tử và mẫu cho 10101 thì
\(\Rightarrow\)A= \(\dfrac{54}{124}\)
\(\Rightarrow\)A= \(\dfrac{27}{62}\)
Vậy A =\(\dfrac{27}{62}\)
à bạn ơi mình đánh nhầm số ở mẫu phải là 10101.(4.31) chứ ko phải 10101.(4.30) đâu
\(\frac{3.5.7.11.13.37+10101}{1212120+40404}=\frac{555555+10101}{1212120+40404}\)
\(=\frac{55.10101+10101}{120.10101+4.10101}\)
\(=\frac{\left(55+1\right).10101}{\left(120+4\right).10101}\)
\(=\frac{56}{124}=\frac{14}{31}\)
\(=\frac{5.11.3.7.11.13-3.7.11.13}{120.10101+4.10101}=\frac{3.7.11.13\left(5.11-1\right)}{10101\left(120+4\right)}=\frac{10101.54}{10101.124}=\frac{54}{124}=\frac{27}{62}\)