Cho B = { m,n }, cho A = { a ; b ; c ; d} Viết tập hợp D sao cho B\(\subset\)D\(\subset\)A
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\(4n-5⋮2n-1\)
\(\Rightarrow4n-2-3⋮2n-1\)
\(\Rightarrow2\left(2n-1\right)-3⋮2n-1\)
\(\Rightarrow3⋮2n-1\)
\(\Rightarrow2n-1\inƯ\left(3\right)\)
\(Ư\left(3\right)=\left\{\pm1;\pm3\right\}\)
\(\Rightarrow\left[{}\begin{matrix}2n-1=1\Rightarrow2n=2\Rightarrow n=1\\2n-1=-1\Rightarrow2n=0\Rightarrow n=0\\2n-1=3\Rightarrow2n=4\Rightarrow n=2\\2n-1=-3\Rightarrow2n=-2\Rightarrow n=-1\end{matrix}\right.\)
2) \(A=\dfrac{9n+7}{3n+4}=\dfrac{9n+12-5}{3n+4}=\dfrac{9n+12}{3n+4}-\dfrac{5}{3n+4}=\dfrac{3\left(3n+4\right)}{3n+4}-\dfrac{5}{3n+4}=3-\dfrac{5}{3n+4}\)
\(\Rightarrow5⋮3n+4\)
\(\Rightarrow3n+4\inƯ\left(5\right)\)
\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(MIN_A\Rightarrow MAX_{3n+4}\)
\(\Rightarrow3n+4=-1\Rightarrow3n=-5\Rightarrow n=-\dfrac{5}{3}\)
Tương tự
\(\Rightarrow\left[{}\begin{matrix}3n+4=1\Rightarrow3n=-3\Rightarrow n=-1\\3n+4=-1\Rightarrow3n=-5\Rightarrow n=-\dfrac{5}{3}\\3n+4=5\Rightarrow3n=1\Rightarrow n=\dfrac{1}{3}\\3n+4=-5\Rightarrow3n=-9\Rightarrow n=-3\end{matrix}\right.\)
b) \(MIN_A\Rightarrow A\in Z^-\Rightarrow3n+4\in Z^-\)
Để \(B\subset D\subset A\)
\(\Rightarrow\) D = {m;n;a;b} = {m;n;a;c}={m;n;b;c}={m;n;b;d}=......
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