ĐKXĐ: \(x\ge0;x\ne4\)
\(P=\left(\dfrac{4}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}-2}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}+\dfrac{\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)\)
\(=\left(\dfrac{4+\sqrt{x}-2+x+2\sqrt{x}}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right).\left(\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\right)\)
\(=\dfrac{\left(x+3\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}.\dfrac{\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+1\right)}\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}\)
b.
\(P=\dfrac{\sqrt{x}-2+4}{\sqrt{x}-2}=1+\dfrac{4}{\sqrt{x}-2}\)
P nguyên lớn nhất khi \(\sqrt{x}-2\) là số nguyên dương nhỏ nhất
\(\Leftrightarrow\sqrt{x}-2=1\Rightarrow\sqrt{x}=3\)
\(\Rightarrow x=9\)
a: ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\notin\left\{4\right\}\end{matrix}\right.\)
\(P=\left(\dfrac{4}{x-4}+\dfrac{1}{\sqrt{x}+2}-\dfrac{\sqrt{x}}{2-\sqrt{x}}\right):\dfrac{\sqrt{x}+1}{\sqrt{x}+2}\)
\(=\left(\dfrac{4}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}+\dfrac{1}{\sqrt{x}+2}+\dfrac{\sqrt{x}}{\sqrt{x}-2}\right)\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(=\dfrac{4+\sqrt{x}-2+\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+2\right)\left(\sqrt{x}-2\right)}\cdot\dfrac{\sqrt{x}+2}{\sqrt{x}+1}\)
\(=\dfrac{\sqrt{x}+2+\left(\sqrt{x}+2\right)\cdot\sqrt{x}}{\sqrt{x}+2}\cdot\dfrac{1}{\sqrt{x}+1}\)
\(=\dfrac{\left(\sqrt{x}+2\right)\left(\sqrt{x}+1\right)}{\left(\sqrt{x}+1\right)\left(\sqrt{x}+2\right)}=1\)
b: Với mọi x nguyên thỏa mãn \(\left\{{}\begin{matrix}x>=0\\x\ne4\end{matrix}\right.\) thì P luôn bằng 1
