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\(P=\sqrt[]{x}+\dfrac{3}{\sqrt[]{x}-1}\left(x>1\right)\)
\(P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\)
Áp dụng bất đẳng thức Cauchy cho 2 số \(\sqrt[]{x}-1;\dfrac{3}{\sqrt[]{x}-1}\) ta được :
\(\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{\sqrt[]{x}-1.\dfrac{3}{\sqrt[]{x}-1}}\)
\(\Rightarrow\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}\ge2\sqrt[]{3}\)
\(\Rightarrow P=\sqrt[]{x}-1+\dfrac{3}{\sqrt[]{x}-1}+1\ge2\sqrt[]{3}+1\)
\(\Rightarrow Min\left(P\right)=2\sqrt[]{3}+1\)
\(Đkxđ\Leftrightarrow\hept{\begin{cases}x>0\\\left(\sqrt{x}-1\right)^2>0\end{cases}\Rightarrow\hept{\begin{cases}x>0\\x>1\end{cases}\Rightarrow}x>1}\)
\(C=\)\(\frac{1}{\sqrt{x}}+\frac{3}{x\sqrt{x}}+1+\frac{2}{x-\sqrt{x}+1}\)
\(=\frac{1}{\sqrt{x}}+\frac{3}{x\sqrt{x}}+1+\frac{2}{\left(\sqrt{x}-1\right)^2}\)
\(=\frac{x\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{3\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{x\sqrt{x}\left(\sqrt{x}-1\right)^2}{x\sqrt{x}\left(\sqrt{x}-1\right)^2}+\frac{2x.\sqrt{x}}{x\sqrt{x}\left(\sqrt{x-1}\right)^2}\)
\(=x\left(\sqrt{x}-1\right)^2+3\left(\sqrt{x}-1\right)^2+x\sqrt{x}\left(\sqrt{x}-1\right)^2+2x.\sqrt{x}\)
.....
ĐKXĐ: x>=0
a: P=1/2
=>\(\dfrac{\sqrt{x}+2}{\sqrt{x}+5}=\dfrac{1}{2}\)
=>\(2\sqrt{x}+4=\sqrt{x}+5\)
=>\(\sqrt{x}=1\)
=>x=1(nhận)
b: \(P^2-P=P\left(P-1\right)\)
\(=\dfrac{\sqrt{x}+2}{\sqrt{x}+5}\cdot\dfrac{\sqrt{x}+2-\sqrt{x}-5}{\sqrt{x}+5}\)
\(=\dfrac{-3\left(\sqrt{x}+2\right)}{\left(\sqrt{x}+5\right)^2}< 0\)
=>\(P^2< P\)
c: Để P nguyên thì \(\sqrt{x}+2⋮\sqrt{x}+5\)
=>\(\sqrt{x}+5-3⋮\sqrt{x}+5\)
=>\(\sqrt{x}+5\inƯ\left(-3\right)\)
=>\(\sqrt{x}+5\in\left\{1;-1;3;-3\right\}\)
=>\(\sqrt{x}\in\left\{-4;-6;-2;-8\right\}\)
=>\(x\in\varnothing\)