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\(\dfrac{1}{3}+\dfrac{3}{35}< \dfrac{x}{210}< \dfrac{4}{7}+\dfrac{1}{3}\)
Ta có: \(\dfrac{1}{3}+\dfrac{3}{35}=\dfrac{35+9}{105}=\dfrac{44}{105}\)
và \(\dfrac{4}{7}+\dfrac{1}{3}=\dfrac{12+7}{21}=\dfrac{19}{21}\)
=> \(\dfrac{44}{105}=\dfrac{44.2}{105.2}=\dfrac{88}{210}\)
=> \(\dfrac{19}{21}=\dfrac{19.10}{21.10}=\dfrac{190}{210}\)
=> \(\dfrac{88}{201}< \dfrac{x}{210}< \dfrac{190}{210}\)
=> Vậy x ∈ {89; 90; 91; 92; ... ; 188; 189}
2: =>3<x<16/5+9/5=5
=>x=4
1: =>70/210+18/210<x/210<120/210+70/210
=>88<x<190
hay \(x\in\left\{89;90;...;189\right\}\)
1. \(\frac{-17}{21}:\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)
\(-\frac{17}{21}:\frac{17}{20}< x+\frac{4}{7}< \frac{7}{12}\)
\(-\frac{20}{21}< x+\frac{4}{7}< \frac{7}{12}\)
\(-\frac{80}{84}< \frac{84x+48}{84}< \frac{49}{84}\)
\(-80< 84x+48< 49\)
\(\begin{cases}-80< 84x+48\\84x+48< 49\end{cases}\)
\(\begin{cases}84x>-128\\84x< 1\end{cases}\)
\(\begin{cases}x>-\frac{32}{21}\\x< \frac{1}{84}\end{cases}\)
\(\Rightarrow-\frac{32}{21}< x< \frac{1}{84}\)
\(-\frac{17}{21}\div\left(\frac{5}{4}-\frac{2}{5}\right)< x+\frac{4}{7}< 1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}\)
\(-\frac{20}{21}< x+\frac{4}{7}< \frac{7}{12}\)
\(-\frac{32}{21}< x< \frac{1}{84}\)
\(-1^{11}_{21}< x< \frac{1}{84}\)
\(\Rightarrow x\in\left\{-1;0\right\}\)
Vậy x = 0
\(\frac{4}{3}\times1,25\times\left(\frac{16}{5}-\frac{5}{16}\right)< 2x< 4-\frac{4}{3}+3-\frac{3}{2}+2\)
\(\frac{77}{16}< 2x< \frac{37}{6}\)
\(\frac{77}{32}< x< \frac{37}{12}\)
\(2^{13}_{32}< x< 3^1_{12}\)
=> x = 3