\(\frac{-25}{35}.\frac{5}{6}< X< \left(-2\right)^2\)
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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 {\(\frac{-874}{189},\frac{-873}{189},......,\frac{107}{189}\)}
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
1)
a)
\(\frac{-5}{6}.\frac{120}{25}< x< \frac{-7}{15}.\frac{9}{14}\)
\(\frac{-1}{1}.\frac{20}{5}< x< \frac{-1}{5}.\frac{3}{2}\)
\(\frac{-20}{5}< x< \frac{-3}{10}\)
\(\frac{-40}{10}< x< \frac{-3}{10}\)
\(\Rightarrow Z\in\left\{-4;-5;-6;-7;-8;-9;-10;...;-39\right\}\)
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) \(\left(\frac{-5}{3}\right)^3< x< \frac{-24}{35}.\frac{-5}{6}\)
\(\Leftrightarrow\frac{-125}{25}< x< \frac{4}{7}\)
\(\Leftrightarrow\frac{-35}{7}< \frac{-7x}{7}< \frac{4}{7}\)
\(\Leftrightarrow-35< -7x< 4\)
\(\Leftrightarrow x\in\left\{4;3;2;1;0\right\}\)
\(\frac{4}{3}.\left(\frac{1}{6}-\frac{1}{2}\right)< x< \frac{2}{3}.\left(\frac{-1}{6}+\frac{3}{4}\right)\)
⇒ \(\frac{4}{3}.\left(\frac{-1}{3}\right)< x< \frac{2}{3}.\left(\frac{7}{12}\right)\)
⇒ \(\frac{-4}{9}< x< \frac{7}{18}\)
⇒ \(\frac{-8}{18}< x< \frac{7}{18}\)
mà -8<x<7
⇒ x ϵ \(\left\{-7;-6;-5;-4;....;5;6\right\}\)
a) ta có \(\frac{-5}{6}\)\(\times\)\(\frac{120}{25}\)< \(x\)<\(\frac{-7}{15}\)\(\times\)\(\frac{4}{9}\)\(\Rightarrow\)\(-4\)<\(x\)<\(-0,2074074074\)\(\Rightarrow\)\(-4\)<\(x\)<\(-0,2\)
mà \(x\)\(\in\)\(ℤ\)\(\Rightarrow\)\(x\)\(\in\)( -1;-2;-3)
b) ta có \(\left(\frac{-5}{3}\right)^3\)<\(x\)<\(\frac{-25}{35}\)\(\times\)\(\frac{-5}{6}\)\(\Rightarrow\)\(-4,62962963\)<\(x\)<\(0,5952380952\)
mà \(x\)\(\in\)\(ℤ\)\(\Rightarrow\)\(x\)\(\in\)(-4;-3;-2;-1;0)
ĐÚNG THÌ K CHO MK NHA