1/5< .../40<1/4 giúp mình với
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B=1/5.(1+1/5+...+\(^{\frac{1}{5^{ }}99}\))
C=1/5.(1/5+\(\frac{1}{5}^3\)+...+\(\frac{1}{5}^{99}\))
- Tính giá trị ở mỗi vế.
- So sánh rồi điền dấu thích hợp vào chỗ trống.
a) 32 + 7 < 40 b) 32 + 14 = 14 + 32
45 + 4 < 54 + 5 69 - 9 < 96 - 6
55 - 5 > 40 + 5 57 - 1 < 57 + 1
Bài 1:
a: \(\Leftrightarrow\dfrac{6}{7}\left(\dfrac{7}{12}x-\dfrac{14}{3}\right)=\dfrac{5}{9}-\dfrac{9}{8}=\dfrac{-41}{72}\)
\(\Leftrightarrow x\cdot\dfrac{7}{12}-\dfrac{14}{3}=-\dfrac{287}{432}\)
\(\Leftrightarrow x\cdot\dfrac{7}{12}=\dfrac{1729}{432}\)
hay \(x=\dfrac{247}{36}\)
b: \(\Leftrightarrow\dfrac{1}{5}:x=\dfrac{1}{5}+\dfrac{3}{14}-\dfrac{8}{7}=\dfrac{-51}{70}\)
hay \(x=-\dfrac{14}{51}\)
c: đề sai rồi bạn
M=\(\dfrac{1919\times171717}{191919\times1717}\) và N=\(\dfrac{18}{19}\)
Ta có :
M= \(\dfrac{1919\times171717}{191919\times1717}\)
M=\(\dfrac{19\times17}{19\times17}\)
M= 1
Mà N= \(\dfrac{18}{19}\)
Vì: 1>\(\dfrac{18}{19}\)
\(\Rightarrow\)\(\dfrac{1919\times171717}{191919\times1717}\) > \(\dfrac{18}{19}\)
\(\Rightarrow\)M > N
A=\(\dfrac{5^{12}+1}{5^{13}+1}\) và B =\(\dfrac{5^{11}+1}{5^{12}+1}\)
Ta có:
A=\(\dfrac{5^{12}+1}{5^{13}+1}\)
\(\Rightarrow\)5.A=5.\(\dfrac{5^{12}+1}{5^{13}+1}\)
=\(\dfrac{5.\left(5^{12}+1\right)}{5^{13}+1}\)
=\(\dfrac{5^{13}+6}{5^{13}+1}\)
=\(\dfrac{\left(5^{13}+1\right)+6}{5^{13}+1}\)
=\(\dfrac{5^{13}+1}{5^{13}+1}\) + \(\dfrac{6}{5^{13}+1}\)
= 1 + \(\dfrac{6}{5^{13}+1}\)
B=\(\dfrac{5^{11}+1}{5^{12}+1}\)
\(\Rightarrow\)5.B = 5.\(\dfrac{5^{11}+1}{5^{12}+1}\)
=\(\dfrac{5.\left(5^{11}+1\right)}{5^{12}+1}\)
=\(\dfrac{5^{12}+6}{5^{12}+1}\)
=\(\dfrac{\left(5^{12}+1\right)+5}{5^{12}+1}\)
=\(\dfrac{5^{12}+1}{5^{12}+1}\) + \(\dfrac{5}{5^{12}+1}\)
= 1 + \(\dfrac{5}{5^{12}+1}\)
Vì: \(5^{13}+1\) > \(5^{12}+1\)
\(\Rightarrow\) \(\dfrac{5}{5^{13}+1}\) < \(\dfrac{5}{5^{12}+1}\)
\(\Rightarrow\) 1+\(\dfrac{5}{5^{13}+1}\) < 1+\(\dfrac{5}{5^{12}+1}\)
\(\Rightarrow\) 5.A < 5.B
\(\Rightarrow\) A < b
`1/5<x/40<1/4`
`8/40 <x/40 < 10/40`
`=> x={9}`
`=> 9/40`
1/5 < 9/40< 1/4