25<52x-1<55
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Ta có \(:\) \(\frac{55}{8}=\frac{55\times9}{8\times9}=\frac{495}{72}\)\(^{\left(1\right)}\)
\(\frac{71}{9}=\frac{71\times8}{9\times8}=\frac{568}{72}\)\(^{\left(2\right)}\)
Từ \(\left(1\right)\)và \(\left(2\right)\)\(\Rightarrow\frac{495}{72}< x< \frac{568}{72}\)
\(\Rightarrow x=\frac{496}{72},\frac{497}{72},\frac{498}{72},...,\frac{567}{72}\)
Vậy \(x=\frac{496}{72},\frac{497}{72},\frac{498}{72},...,\frac{567}{72}\)
\(1)\frac{1}{5}+\frac{2}{11}< \frac{x}{55}< \frac{2}{5}+\frac{1}{55}\)
\(\Rightarrow\frac{11}{55}+\frac{10}{55}< \frac{x}{55}< \frac{22}{55}+\frac{1}{55}\)
\(\Rightarrow\frac{21}{55}< \frac{x}{55}< \frac{23}{55}\)
\(\Rightarrow21< x< 23\)
\(\Rightarrow x=22\)
\(2)\frac{11}{3}+\frac{-19}{6}+\frac{-15}{2}\le x\le\frac{19}{12}+\frac{-5}{4}+\frac{-10}{3}\)
\(\Rightarrow\frac{22}{6}+\frac{-19}{6}+\frac{-45}{6}\le x\le\frac{19}{12}+\frac{-15}{12}+\frac{-40}{12}\)
\(\Rightarrow\frac{22+\left[-19\right]+\left[-45\right]}{6}\le x\le\frac{19+\left[-15\right]+\left[-40\right]}{12}\)
\(=\frac{-42}{6}\le x\le\frac{-36}{12}\)
\(\Rightarrow-7\le x\le-3\)
\(\Rightarrow x\in\left\{-7;-6;-5;-4;-3\right\}\)
52 \(\times\) ( y : 78) = 3380
y : 78 = 3380: 52
y : 78 = 65
y = 65 \(\times\) 78
y = 5070
y: 55 - y + 33 = 76
\(y\) \(\times\) ( \(\dfrac{1}{55}\) - 1) = 76 - 33
y \(\times\) ( \(\dfrac{-54}{55}\) ) = 43
y = 43 : (- \(\dfrac{54}{55}\))
y = - \(\dfrac{2365}{54}\)
= 11/55+ 10/55<x/55<2/5+1/55
=2/5<x/55<23/55
=22/55<x/55<23/55
=> x thuộc rỗng
x/21=4/7+(-7)/3
= x/21=12/21+(-49)/21
= x/21=-37/21
=> x = -37
\(25< 5^{2x-1}< 5^5\)
\(5^2< 5^{2x-1}< 5^5\)
\(2< 2x-1< 5\)
\(2+1< 2x< 5+1\)
\(3< 2x< 6\)
\(\dfrac{3}{2}< x< 3\)
x=2