\(\frac{11}{2}-2,4< a< 1,35:0,15\left(a\in N\right)\)
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18 tháng 12 2016
c) \(\frac{0,375-0,3+\frac{3}{11}+\frac{3}{12}}{0,625-0,5+\frac{5}{11}+\frac{5}{12}}=\frac{3\left(0,125-0,1+\frac{1}{11}+\frac{1}{12}\right)}{5\left(0,123-0,1+\frac{1}{11}+\frac{1}{12}\right)}=\frac{3}{5}\)
BA
18 tháng 11 2019
Ta có
\(A=\frac{1}{14}+\frac{1}{29}+...+\frac{1}{n^2+\left(n+1\right)^2+\left(n+2\right)^2}+...+\frac{1}{1877}\)
\(=\frac{1}{1^2+2^2+3^2}+\frac{1}{2^2+3^2+4^2}+...+\frac{1}{n^2+\left(n+1\right)^2+\left(n+2\right)^2}+...+\frac{1}{24^2+25^2+26^2}\)
\(B=n^2+\left(n+1\right)^2+\left(n+2\right)^2=3n^2+6n+5\left(1\right)\)
+ Với \(n\ge1\)từ (1) ta có \(B\le3n^2+9n+6=3\left(n^2+3n+2\right)=3\left(n+1\right)\left(n+2\right)\)Từ đó
\(A>\frac{1}{3}\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\right)=\frac{1}{3}C\)
Với \(C=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}=\frac{1}{2}-\frac{1}{26}=\frac{6}{13}\)
\(\Rightarrow A>\frac{1}{3}\cdot\frac{6}{13}=\frac{2}{13}>0,15\)
+ Với \(n\ge1\)từ (1) ta có \(B>2n^2+6n+4=2\left(n^2+3n+2\right)=2\left(n+1\right)\left(n+2\right)\)
\(\Rightarrow A< \frac{1}{2}\left(\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\right)=\frac{1}{2}C\)
Với \(C=\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+....+\frac{1}{\left(n+1\right)\left(n+2\right)}+...+\frac{1}{24\cdot25}+\frac{1}{25\cdot26}\)
\(=\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{25}-\frac{1}{26}=\frac{1}{2}-\frac{1}{26}=\frac{6}{13}\)
\(\Rightarrow A< \frac{1}{2}\cdot\frac{6}{13}=\frac{3}{13}< 0,25\)
Vậy \(0,15< A< 0,25\)
VM
26 tháng 2 2020
Bài 1:
a) \(0,5-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)
\(=\frac{1}{2}-\frac{5}{41}+\frac{1}{2}-\frac{36}{41}\)
\(=\left(\frac{1}{2}+\frac{1}{2}\right)-\left(\frac{5}{41}+\frac{36}{41}\right)\)
\(=1-1\)
\(=0.\)
b) \(\left(-\frac{2}{3}+\frac{3}{7}\right):\frac{4}{5}+\left(-\frac{1}{3}+\frac{4}{7}\right):\frac{4}{5}\)
\(=-\frac{2}{3}+\frac{3}{7}:\frac{4}{5}-\frac{1}{3}+\frac{4}{7}:\frac{4}{5}\)
\(=\left[\left(-\frac{2}{3}\right)-\frac{1}{3}\right]+\left(\frac{3}{7}+\frac{4}{7}\right):\frac{4}{5}\)
\(=\left(-1\right)+1:\frac{4}{5}\)
\(=\left(-1\right)+\frac{5}{4}\)
\(=\frac{1}{4}.\)
c) \(\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.\sqrt{49}}\)
\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+3.7}\)
\(=\left(-\frac{3}{4}\right).\sqrt{\frac{16}{9}+21}\)
\(=\left(-\frac{3}{4}\right).\sqrt{\frac{205}{9}}\)
\(=\left(-\frac{3}{4}\right).\frac{\sqrt{205}}{3}\)
\(=-\frac{\sqrt{205}}{4}.\)
d) \(\left(-\frac{1}{3}\right)^2.\frac{4}{11}+1\frac{5}{11}.\left(\frac{1}{3}\right)^2\)
\(=\frac{1}{9}.\frac{4}{11}+\frac{16}{11}.\frac{1}{9}\)
\(=\frac{1}{9}.\left(\frac{4}{11}+\frac{16}{11}\right)\)
\(=\frac{1}{9}.\frac{20}{11}\)
\(=\frac{20}{99}.\)
Chúc bạn học tốt!
CV
11 tháng 12 2017
A=[(1+2+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x (2,4x42 - 21x4,8)] / 1+1/2+1/3+...+1/100
= [(1+2+3+...+100) x (1/2 - 1/3 - 1/4-1/5) x (2,4x2x21 - 21x2x 4,8)] / 1+1/2+1/3+...+1/100
=[(1+2+3+...+100) x (1/2 - 1/3 - 1/4 - 1/5) x 0] / 1+1/2+1/3+...+1/100
=0 / 1+1/2+1/3+...+1/100 = 0
\(\frac{11}{2}-2,4< a< 1,35:0,15\)
=> 3,1 < a < 9
MÀ a thuộc N
=> a \(\in\){ 4 ; 5 ; 6 ; 7 ; 8 }
Vậy a \(\in\){ 4 ; 5 ; 6 ; 7 ; 8 }
Ta có :
11/2 - 12/5 < a < 27/20 . 20/3
3,1 < a < 9
Vì a thuộc N nên :
---> a thuộc { 4 ; 5 ; 6 ; 7 ; 8 }