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a: =>19/23>19/x>19/29
=>\(x\in\left\{24;25;26;27;28\right\}\)
b: =>88/132<88/x<88/128
=>132>x>128
=>\(x\in\left\{131;130;129\right\}\)
c: =>\(\left\{{}\begin{matrix}\dfrac{4}{x}-\dfrac{x}{8}< 0\\\dfrac{x}{8}-\dfrac{5}{x}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}\dfrac{32-x^2}{8x}< 0\\\dfrac{x^2-40}{8x}< 0\end{matrix}\right.\)
=>32<x^2<40
=>x=6
\(a)-\frac{5}{19}+\frac{3}{19}< \frac{x}{19}< \frac{13}{19}+-\frac{11}{19}\)
\(\Rightarrow\frac{-5+3}{19}< \frac{x}{19}< \frac{13+(-11)}{19}\)
\(\Rightarrow-\frac{2}{19}< \frac{x}{19}< \frac{2}{19}\)
Tìm x đi rồi ra được kết quả
`(2/(11.13)+2/(13.15)+....+2/(19.21)).462-[0,04/(x+1,05):0,12=19`
`=>(1/11-1/13+1/13-1/15+....+1/19-1/21).462-(2/(50(x+1,05)).25/3=19`
`=>(1/11-1/21).462-1/(3(x+1,05))=19`
`=>10/231. 462-1/(3x+3,15)=19`
`=>20-1/(3x+3,15)=19
`=>1/(3x+3,15)=1`
`=>3x+3,15=1`
`=>3x=-2,15`
`=>x=-43/60`
Vậy `x=-43/60`
1:
a: \(=\dfrac{-4}{7}+\dfrac{4}{7}+\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=\dfrac{3}{7}-\dfrac{23}{34}-\dfrac{4}{5}=-\dfrac{1247}{1190}\)
b:
Sửa đề: \(\dfrac{-5}{13}+\dfrac{4}{19}+\dfrac{-8}{13}+\dfrac{15}{19}+\dfrac{45}{6}\)
\(=\dfrac{-5}{13}-\dfrac{8}{13}+\dfrac{4}{19}+\dfrac{15}{19}+\dfrac{45}{6}=\dfrac{9}{2}\)
A=\(\frac{6}{19}\). \(\frac{-7}{11}\)+\(\frac{6}{19}\).\(\frac{-4}{11}\)+\(\frac{-13}{19}\)
=\(\frac{6}{19}\).(\(\frac{-7}{11}\)+\(\frac{-4}{11}\))+\(\frac{-13}{19}\)
=\(\frac{6}{19}\).\(\frac{-11}{11}\)+\(\frac{-13}{19}\)
=\(\frac{6}{19}\).-1 +\(\frac{-13}{19}\)
=\(\frac{-6}{19}\)+\(\frac{-13}{19}\)
=\(\frac{-19}{19}\)
+1
- 5 19 + 3 19 < x 19 ≤ 13 19 + - 11 19 ⇔ - 2 19 < x 19 ≤ 2 19 - 2 < x ≤ 2
Mà x thuộc Z
Do đó x ∈ - 1 ; 0 ; 1 ; 2