Tìm x thuộc Z, biết: \(1+\frac{-1}{60}+\frac{19}{120}
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\(1+\frac{-1}{60}+\frac{19}{120}< \frac{x}{36}< \frac{58}{90}+\frac{59}{72}+\frac{-1}{60}\)
=> \(\frac{137}{120}< \frac{x}{36}< \frac{521}{360}\)
=> \(\frac{411}{360}< \frac{10x}{360}< \frac{521}{360}\)
=> 411 < 10x < 521
=> x \(\in\){ 42,43,44,...,52}
Bài 1 :
\(A=\dfrac{n+1}{n+2}\) có giá trị nguyên âm, dương khi
\(n+1⋮n+2\)
\(\Rightarrow n+1-\left(n+2\right)⋮n+2\)
\(\Rightarrow n+1-n-2⋮n+2\)
\(\Rightarrow-1⋮n+2\)
\(\Rightarrow n+2\in\left\{-1;1\right\}\)
\(\Rightarrow n\in\left\{-3;-1\right\}\left(n\in Z\right)\)
Bài 2 :
\(1+\left(-\dfrac{1}{60}\right)+\dfrac{19}{120}< \dfrac{x}{36}+\left(-\dfrac{1}{60}\right)< \dfrac{58}{90}+\dfrac{59}{72}+\left(-\dfrac{1}{60}\right)\)
\(\Rightarrow1+\dfrac{19}{120}< \dfrac{x}{36}< \dfrac{58}{90}+\dfrac{59}{72}\)
\(\Rightarrow\dfrac{139}{120}< \dfrac{x}{36}< \dfrac{232}{360}+\dfrac{295}{360}\)
\(\Rightarrow\dfrac{417}{360}< \dfrac{10x}{360}< \dfrac{527}{360}\)
\(\Rightarrow417< 10x< 527\)
\(\Rightarrow10x\in\left\{420;430;440;450;460;470;480;490;500;510;520\right\}\)
\(\Rightarrow x\in\left\{42;43;44;45;46;47;48;49;50;51;52\right\}\)
\(1+\frac{-1}{60}+\frac{19}{120}