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a. Vì A thuộc Z
\(\Rightarrow x-2\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x\in\left\{-3;1;3;7\right\}\)( tm x thuộc Z )
b. Ta có : \(B=\frac{x+2}{x-3}=\frac{x-3+5}{x-3}=1+\frac{5}{x-3}\)
Vì B thuộc Z nên 5 / x - 3 thuộc Z
\(\Rightarrow x-3\in\left\{-5;-1;1;5\right\}\)
\(\Rightarrow x\in\left\{-2;2;4;8\right\}\)( tm x thuộc Z )
c. Ta có : \(C=\frac{x^2-x}{x+1}=\frac{x^2+x-2x+2-2}{x+1}=\frac{x\left(x+1\right)-2x+2-2}{x+1}\)
\(=x-2-\frac{2}{x+1}\)
Vi C thuộc Z nên 2 / x + 1 thuộc Z
\(\Rightarrow x+1\in\left\{-2;-1;1;2\right\}\)
\(\Rightarrow x\in\left\{-3;-2;0;1\right\}\) ( tm x thuộc Z )
Bài 3 :
c) \(\dfrac{m}{5}-\dfrac{2}{n}=\dfrac{2}{5}\)
\(\Leftrightarrow\) \(\dfrac{m}{5}-\dfrac{2}{5}=\dfrac{2}{n}\)
\(\Leftrightarrow\) \(\dfrac{m-2}{5}=\dfrac{2}{n}\)
\(\Rightarrow\) ( m - 2 ) . n = 10
10 có các ước là : \(\pm1;\pm2;\pm5;\pm10\)
*\(\left\{{}\begin{matrix}m-2=1\\n=10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=3\\n=10\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}m-2=-1\\n=-10\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=1\\n=-10\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}m-2=10\\n=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=12\\n=1\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}m-2=-10\\n=-1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=-8\\n=-1\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}m-2=2\\n=5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=4\\n=5\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}m-2=-2\\n=-5\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=0\\n=-5\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}m-2=5\\n=2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=7\\n=2\end{matrix}\right.\)
*\(\left\{{}\begin{matrix}m-2=-5\\n=-2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}m=-3\\n=-2\end{matrix}\right.\)
Vậy có 8 cặp (m,n) thỏa mãn : (3,10) ; (1,-10) ; (12,1) ; (-8,-1) ; (4,5) ; (0,-5) ; (7,2) ; (-3,-2) .
Bài 5:Giải:
Ta có: \(\left\{{}\begin{matrix}a+3c=2016\left(1\right)\\a+2b=2017\left(2\right)\end{matrix}\right.\)
Từ \(\left(1\right)\Leftrightarrow a=2016-3c\)
Lấy \(\left(2\right)-\left(1\right)\) ta được:
\(2b-3c=1\Leftrightarrow b=\dfrac{1+3c}{2}\)
Khi đó:
\(P=a+b+c=\left(2016-3c\right)+\dfrac{1+3c}{2}\) \(+\) \(c\)
\(=\left(2016+\dfrac{1}{2}\right)+\dfrac{-6c+3c+2c}{2}\)
\(=2016\dfrac{1}{2}-\dfrac{c}{2}\) Vì \(a,b,c\ge0\) nên:
\(P=2016\dfrac{1}{2}-\dfrac{c}{2}\le2016\dfrac{1}{2}\)
Vậy \(P_{max}=2016\dfrac{1}{2}\Leftrightarrow c=0\)