tìm x
\(\frac{-37}{8}+\frac{13}{8}< x< \frac{1}{4}+\left(-\frac{5}{4}\right)\) )
K
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EC
8 tháng 9 2016
\(\left(-\frac{2}{3}:-\frac{1}{3}\right).\left(-\frac{9}{2}\right)-\frac{1}{4}< \frac{x}{8}< -\frac{1}{2}.\frac{3}{4}:\frac{1}{8}+1\)
\(2.\left(-\frac{9}{2}\right)-\frac{1}{4}< \frac{x}{8}< \left(-3\right)+1\)
\(\left(-9\right)-\frac{1}{4}< \frac{x}{8}< \left(-2\right)\)
\(\left(-\frac{37}{4}\right)< \frac{x}{8}< \left(-2\right)\)
\(-\frac{74}{8}< \frac{x}{8}< -\frac{16}{8}\)
Vậy -74<x<-16
KT
11 tháng 8 2018
\(\frac{1}{4}+\frac{8}{9}\le\frac{x}{36}\le1-\left(\frac{3}{8}-\frac{5}{6}\right)\)
<=> \(\frac{41}{36}\le\frac{x}{36}\le\frac{35}{24}\)
<=> \(\frac{82}{72}\le\frac{2x}{72}\le\frac{105}{72}\)
<=> \(82\le2x\le105\)
<=> \(41\le x\le52,5\)
Do \(x\in N\)nên \(x=\left\{x\in N|41\le x\le52,5\right\}\)
9 tháng 8 2019
\(a,\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)< x< \left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}\)
\(taco:\left(\frac{31}{20}-\frac{26}{45}\right)\cdot\left(\frac{-36}{35}\right)=\frac{35}{36}\cdot\frac{-36}{35}=-1\)
\(\left(\frac{51}{56}+\frac{8}{21}+\frac{1}{3}\right)\cdot\frac{8}{13}=\frac{13}{8}\cdot\frac{8}{13}=1\)
\(=>x=0\)
\(b,\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}< x< \frac{-1}{2}+2+\frac{5}{2}\)(dau <co dau gach ngang o duoi nha)
\(taco:\frac{-5}{6}+\frac{8}{3}+\frac{29}{-3}=\frac{-5}{6}+\frac{8}{3}+\frac{-29}{3}=\frac{-5}{6}+\frac{16}{6}+\frac{-58}{6}=\frac{-47}{6}=-7,8\)
\(\frac{-1}{2}+2+\frac{5}{2}=\frac{3}{2}+\frac{5}{2}=4\)
tu do \(=>x=-7,8;...;0;1;2;3;4\)
\(\frac{-37}{8}+\frac{13}{8}< x< \frac{1}{4}+\left(-\frac{5}{4}\right)\)
\(\frac{-37+13}{8}< x< \frac{1-5}{4}\)
\(\frac{-24}{8}< x< \frac{-4}{4}\)
\(\Leftrightarrow-3< x< -1\)
=> x = -2