Bài 6:
\(a,B>0\Leftrightarrow\dfrac{x^2-x+\dfrac{1}{4}+\dfrac{7}{4}}{x-3}>0\Leftrightarrow\dfrac{\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}{x-3}>0\\ \Leftrightarrow x-3>0\left[\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>0\right]\\ \Leftrightarrow x>3\\ b,B\in Z\Leftrightarrow\dfrac{x\left(x-3\right)+2\left(x-3\right)+8}{x-3}=x+2+\dfrac{8}{x-3}\in Z\\ \Leftrightarrow x-3\inƯ\left(8\right)=\left\{-8;-4;-2;-1;1;2;4;8\right\}\\ \Leftrightarrow x\in\left\{-5;-1;1;2;4;5;7;11\right\}\)
Bài 7:
\(a,M=\dfrac{8}{x^2-4x+4+8}=\dfrac{8}{\left(x-2\right)^2+8}\le\dfrac{8}{0+8}=1\\ M_{max}=1\Leftrightarrow x=2\)
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