tìm \(a\inℤ\)biết: \(\frac{1}{8}< \frac{3}{a}< \frac{1}{7}\)
K
Khách
Hãy nhập câu hỏi của bạn vào đây, nếu là tài khoản VIP, bạn sẽ được ưu tiên trả lời.
Những câu hỏi liên quan
BH
16 tháng 4 2019
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
=>a
23 tháng 7 2018
a) \(\frac{1}{3}+\frac{3}{35}< \frac{x}{210}< \frac{7}{7}+\frac{3}{5}+\frac{1}{3}\)
\(\Rightarrow\frac{44}{105}< \frac{x}{210}< \frac{29}{15}\)
\(\Rightarrow\frac{88}{210}< \frac{x}{210}< \frac{406}{210}\)
\(\Rightarrow x\in\left\{89;90;91;...;405\right\}\)
b) \(\frac{5}{3}+-\frac{14}{3}< x< \frac{8}{5}+\frac{8}{10}\)
\(\Rightarrow-3< x< 2\frac{2}{5}\)
=> x thuộc {-2;-1;0;1;2} ( nếu x là số nguyên)
11 tháng 4 2019
\(a,\left[\frac{4}{5}+\frac{2}{3}\right]:\frac{1}{5}-1,4\cdot\left[\frac{-5}{7}\right]^2\)
\(=\left[\frac{4\cdot3}{15}+\frac{2\cdot5}{15}\right]:\frac{1}{5}-1,4\cdot\frac{-5}{7}\cdot\frac{-5}{7}\)
\(=\left[\frac{12}{15}+\frac{10}{15}\right]:\frac{1}{5}-\frac{14}{10}\cdot\frac{25}{49}\)
\(=\frac{22}{15}:\frac{1}{5}-\frac{7}{5}\cdot\frac{25}{49}\)
\(=\frac{22}{15}\cdot\frac{5}{1}-\frac{7}{5}\cdot\frac{25}{49}\)
\(=\frac{22\cdot5}{15\cdot1}-\frac{7\cdot25}{5\cdot49}=\frac{22\cdot1}{3\cdot1}-\frac{1\cdot5}{1\cdot7}=\frac{22}{3}-\frac{5}{7}\)
= ...
Tự tính
Bài 2 : \(a,3-\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{2}\)
\(\Rightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{1}{2}+3\)
\(\Rightarrow\left|\frac{1}{2}x-\frac{1}{3}\right|=\frac{7}{2}\)
\(\Rightarrow\hept{\begin{cases}\frac{1}{2}x-\frac{1}{3}=\frac{7}{2}\\\frac{1}{2}x-\frac{1}{3}=-\frac{7}{2}\end{cases}}\Rightarrow\hept{\begin{cases}x=\frac{23}{3}\\x=\frac{-19}{3}\end{cases}}\)
Vậy \(x\in\left\{\frac{23}{3};\frac{-19}{3}\right\}\)
b, \(0,6-160\%< x\le3\frac{2}{3}:\frac{22}{18}\)
\(\Rightarrow0,6-\frac{160}{100}< x\le\frac{11}{3}:\frac{22}{18}\)
\(\Rightarrow0,6-\frac{8}{5}< x\le\frac{11}{3}\cdot\frac{18}{22}\)
\(\Rightarrow0,6-1,6< x\le3\)
\(\Rightarrow-1< x\le3\)
\(\Rightarrow x\in\left\{0;1;2;3\right\}\)
\(~~~hd~~~\)
\(\frac{1}{8}< \frac{3}{a}< \frac{1}{7}\Leftrightarrow\frac{3}{24}< \frac{3}{a}< \frac{3}{21}\Leftrightarrow24< a< 21\Leftrightarrow a\in\left\{22;23\right\}\)
ta có: \(\frac{1}{8}< \frac{3}{a}< \frac{1}{7}\)
\(\Rightarrow\frac{3}{24}< \frac{3}{a}< \frac{3}{21}\)
\(\Rightarrow a\in\left\{22;23\right\}\)
vậy; a= 22; hoặc a= 23