a) điều kiện \(x\in Q\)
ta có : \(A=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3+5}{\sqrt{x}+3}=1+\dfrac{5}{\sqrt{x}+3}\)
ta có : \(A\) nguyên \(\Leftrightarrow\) \(\dfrac{5}{\sqrt{x}+3}\) nguyên \(\Leftrightarrow\) \(\sqrt{x}+3\) thuộc ước của 5 là \(\pm1;\pm5\)
ta có : * \(\sqrt{x}+3=1\Leftrightarrow\sqrt{x}=-2\left(vôlí\right)\)
* \(\sqrt{x}+3=-1\Leftrightarrow\sqrt{x}=-4\left(vôlí\right)\)
* \(\sqrt{x}+3=5\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tmđk\right)\)
* \(\sqrt{x}+3=-5\Leftrightarrow\sqrt{x}=-8\left(vôlí\right)\)
mình sửa đề tí mẫu là căn(x) + 8
sr nha mn
\(A=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3+5}{\sqrt{x}+3}=\dfrac{\sqrt{x}+3}{\sqrt{x}+3}+\dfrac{5}{\sqrt{x+3}}\)
\(\Rightarrow5⋮\sqrt{x+3}\)
\(\Rightarrow\sqrt{x}+3\inƯ\left(5\right)\)
\(Ư\left(5\right)=\left\{\pm1;\pm5\right\}\)
\(\Rightarrow\left\{{}\begin{matrix}\sqrt{x}+3=1\Rightarrow\sqrt{x}=-2\left(KTM\right)\\\sqrt{x}+3=-1\Rightarrow\sqrt{x}=-4\left(KTM\right)\\\sqrt{x}+3=5\Rightarrow\sqrt{x}=2\Rightarrow x=4\left(TM\right)\\\sqrt{x}+3=-5\Rightarrow\sqrt{x}=-8\end{matrix}\right.\)
Vậy \(x=4\)
\(B=\sqrt{x}.A\)
\(B=\sqrt{4}.A\)
\(B=-2.A\) ( vì -2 nhỏ hơn A)
\(\Rightarrow B=-2\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)
\(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\)
\(\Rightarrow MIN_A=\dfrac{\sqrt{0}+8}{\sqrt{0}+3}=\dfrac{8}{3}\)
Vậy \(MIN_B=-2.\dfrac{8}{3}=\dfrac{-16}{3}\)
