Tìm \(a\inℤ\), biết \(\frac{-2}{3}+\frac{1}{4}< \frac{a}{6}< \frac{3}{4}-\frac{1}{3}\)
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1 tháng 5 2019
a) Đặt \(B=\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{2014^2}\)
Ta có: \(\frac{1}{2^2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}< \frac{1}{2.3}\)
.................
\(\frac{1}{2014^2}< \frac{1}{2013.2014}\)
\(\Rightarrow B< \frac{1}{1.2}+\frac{1}{2.3}+...+\frac{1}{2013.2014}\)
\(\Rightarrow B< 1-\frac{1}{2014}< 1\)
\(\Rightarrow B< 1\)
\(\Rightarrow1+B< 1+1\)
Hay \(A< 2\)
1 tháng 5 2019
C) Ta có: \(\frac{1}{2}< \frac{2}{3}\)
\(\frac{3}{4}< \frac{4}{5}\)
.................
\(\frac{9999}{10000}< \frac{10000}{10001}\)
\(\Rightarrow C< \frac{2}{3}.\frac{4}{5}.....\frac{10000}{10001}\)
\(\Rightarrow C^2< \left(\frac{1}{2}.\frac{3}{4}.....\frac{9999}{10000}\right).\left(\frac{2}{3}.\frac{4}{5}.....\frac{10000}{10001}\right)\)
\(\Rightarrow C^2< \frac{1}{10001}< \frac{1}{10000}\)
\(\Rightarrow C^2< \frac{1}{10000}\)
\(\Rightarrow C< \frac{1}{100}\)
30 tháng 4 2018
A=\(1+\left(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}\right)\)
Đặt B=\(\frac{1}{2^2}+\frac{1}{3^2}+...+\frac{1}{100^2}< \frac{1}{1.2}+\frac{1}{2.3}+..+\)\(\frac{1}{99.100}=\)\(1-\frac{1}{100}< 1\)
Mà A=1+B=>A=1+B<1+1=2
30 tháng 4 2018
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< 2\)
\(A=1+\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}< \frac{1}{1\cdot2}+\frac{1}{2\cdot3}+\frac{1}{3\cdot4}+...+\frac{1}{99\cdot100}\)
\(A=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A=1-\frac{1}{100}\)
vậy \(A=\frac{99}{100}< 2\left(đpcm\right)\)
B)
ta có : \(1=1\)
\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)
\(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{7}< \frac{1}{4}+...+\frac{1}{4}=\frac{4}{4}=1\)
\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}< \frac{1}{8}+...+\frac{1}{8}=\frac{8}{8}=1\)
\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{63}< 1\)
tất cả công lại \(\Rightarrow B< 6\)
25 tháng 8 2018
Bài 1 : Thực hiện phép tính :
a, \(\frac{4}{5}+1\frac{1}{6}\cdot\frac{3}{4}\)
= \(\frac{4}{5}+\frac{7}{6}\cdot\frac{3}{4}\)
= \(\frac{4}{5}+\frac{7}{8}\)
= \(\frac{32+35}{40}=\frac{67}{40}\)
b, \(\frac{2}{3}:\left(\frac{3}{4}\cdot\frac{4}{3}\right)+2\)
\(=\frac{2}{3}:1+2\)
\(=\frac{2}{3}+2=\frac{2+6}{3}=\frac{8}{3}\)
c, \(\frac{1}{2}\times\left(\frac{2}{3}+\frac{3}{5}\cdot\frac{5}{7}\right)+1\frac{1}{3}\)
\(=\frac{1}{2}\cdot\left(\frac{2}{3}+\frac{9}{35}\right)+\frac{4}{3}\)
\(=\frac{1}{2}\cdot\frac{97}{105}+\frac{4}{3}\)
\(=\frac{97}{210}+\frac{4}{3}=\frac{377}{210}\)
Bài 2 : Tìm \(x\inℤ\), biết :
a, \(\frac{2}{3}< \frac{x}{6}\le\frac{10}{3}\)
\(\Leftrightarrow\frac{4}{6}< \frac{x}{6}\le\frac{20}{6}\)
mà \(x\inℤ\Rightarrow\text{x}\in\) {\(5;6;7;8;9;10;11;12;13;14;15;16;17;18;19;20\)}
b, \(\frac{1}{3}+x=1\frac{1}{2}\)
\(\frac{1}{3}+x=\frac{3}{2}\)
\(x=\frac{3}{2}+\frac{\left(-1\right)}{3}\)
\(x=\frac{7}{6}\) (loại vì \(x\notinℤ\))
\(\Rightarrow x\in\varnothing\)
c, \(\frac{1}{7}+x=\frac{25}{14}+\frac{5}{14}\)
\(\frac{1}{7}+x=\frac{15}{7}\)
\(x=\frac{15}{7}+\frac{(-1)}{7}\)
\(x=\frac{14}{7}=2\).
Ta có\(\frac{-2}{3}\)+\(\frac{1}{4}\)= \(\frac{-8}{12}\)+\(\frac{3}{12}\)= \(\frac{-5}{12}\)
\(\frac{3}{4}\)-\(\frac{1}{3}\)=\(\frac{9}{12}\)-\(\frac{4}{12}\)=\(\frac{5}{12}\)
=> \(\frac{-5}{12}\)<\(\frac{a}{6}\)<\(\frac{5}{12}\)
=> \(\frac{-5}{12}\)<\(\frac{2a}{12}\)<\(\frac{5}{12}\)
Mà a là số nguyên,2a là số chẵn
=>2a
{-4,-2,0,2,4}
=>a
{-2,-1,0,1,2}