Tìm x biết \(\left(-\frac{5}{3}\right)^2< x< -\frac{24}{35}\cdot-\frac{5}{6}\)
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Ta có:
\(\left(\frac{-5}{3}\right)^2=\frac{25}{15}=\frac{5}{3}\)
\(\frac{-24}{35}\cdot\frac{-5}{6}=\frac{120}{210}=\frac{4}{7}\)
Quy đồng \(\frac{5}{3}\)và \(\frac{4}{7}\),ta được:
\(\frac{35}{21}\)và \(\frac{12}{21}\)
Vì 35 > 12 nên \(\frac{5}{3}>\frac{4}{7}\)
mà x lại lớn hơn \(\frac{5}{3}\)và bé hơn \(\frac{4}{7}\)
\(\Rightarrow\)Không tồn tại x
a)\(\frac{-5}{6}\).\(\frac{120}{25}\)<x<\(\frac{-7}{15}\).\(\frac{9}{14}\)
-4 <x<\(\frac{-3}{10}\)
\(\frac{-40}{10}\)< x <\(\frac{-3}{10}\)=>x E {-39:-38:-37:.....:-4}
b)\(\left(\frac{-5}{3}\right)^3\)<x<\(\frac{-24}{35}.\frac{-5}{6}\)
\(\frac{-875}{189}< x< \frac{108}{189}\)
=> x E {\(\frac{-874}{189},\frac{-873}{189},......,\frac{107}{189}\)}
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
a)Ta có: 1/2-(1/3+1/4)= -1/12
1/48-(1/16-1/6)=1/8
suy ra: -1/12<x<1/8
<=> -2/24<x<3/24
=>x thuộc:(-1/24 ;0 ;1/24 ;2/24 ;3/24)
1) \(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\Leftrightarrow\frac{-5}{6}.\frac{24}{5}< x< \frac{-63}{210}\)
\(\Leftrightarrow-40< x< \frac{-63}{210}\)
\(\Leftrightarrow\frac{-400}{10}< \frac{10x}{10}< \frac{-3}{10}\)
\(\Leftrightarrow-400< 10x< -3\)
\(\Leftrightarrow x\in\left\{-39;-38;...;-2;-1\right\}\)
2) \(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)
\(\Leftrightarrow\frac{-125}{25}< x< \frac{4}{7}\)
\(\Leftrightarrow\frac{-35}{7}< \frac{-7x}{7}< \frac{4}{7}\)
\(\Leftrightarrow-35< -7x< 4\)
\(\Leftrightarrow x\in\left\{4;3;2;1;0\right\}\)