Bài 1:
Ta có:
\(P=\dfrac{\sqrt{x} +7}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)+6}{\sqrt{x}+1}=1+\dfrac{6}{\sqrt{x}+1}\left(x\ge0\right)\)
Để P nguyên thì: \(\sqrt{x}+1\inƯ\left(6\right)=\left\{1;-1;2;-2;3;-3;6;-6\right\}\)
Mà: \(\sqrt{x}+1\ge1\forall\left(x\ge0\right)\)
\(\Rightarrow\sqrt{x}+1\in\left\{1;2;3;6\right\}\\ \Rightarrow\sqrt{x}\in\left\{0;1;2;5\right\}\\ \Rightarrow x\in\left\{0;1;4;25\right\}\)
Bài 2:
\(A=\dfrac{n+1}{2n-3}=\dfrac{1}{2}\cdot\dfrac{2n+2}{2n-3}=\dfrac{1}{2}\left(\dfrac{2n-3+5}{2n-3}\right)=\dfrac{1}{2}\cdot\left(1+\dfrac{5}{2n-3}\right)\)
Để A nhỏ nhất thì \(1+\dfrac{5}{2n-3}\) nhỏ nhất
=>2n-3=-1
=>2n=2
=>n=1
Bài 1:
ĐKXĐ: x>=0
Để P nguyên thì \(\sqrt{x}+7⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1+6⋮\sqrt{x}+1\)
=>\(6⋮\sqrt{x}+1\)
=>\(\sqrt{x}+1\in\left\{1;2;3;6\right\}\)
=>\(\sqrt{x}\in\left\{0;1;2;5\right\}\)
=>\(x\in\left\{0;1;4;25\right\}\)
