\(Q=\dfrac{\sqrt{x}}{x+\sqrt{x}+1}\Rightarrow\dfrac{1}{Q}=\dfrac{x+\sqrt{x}+1}{\sqrt{x}}\)

\(=\dfrac{x}{\sqrt{x}}+\dfrac{\sqrt{x}}{\sqrt{x}}+\dfrac{1}{\sqrt{x}}=\sqrt{x}+1+\dfrac{1}{\sqrt{x}}\)

\(\ge2\sqrt{\sqrt{x}\cdot\dfrac{1}{\sqrt{x}}}+1=2+1=3\)

Khi x=1

Ko hiểu sao lại cho $$x\ne 0$$

\(\left\{{}\begin{matrix}a\left(a+b+c\right)=36\\b\left(a+b+c\right)=48\\c\left(a+b+c\right)=60\end{matrix}\right.\)

Cộng theo vế 3 pt trên ta có:

\(a\left(a+b+c\right)+b\left(a+b+c\right)+c\left(a+b+c\right)=144\)

\(\Leftrightarrow\left(a+b+c\right)^2=144\Leftrightarrow a+b+c=12\)

\(\Rightarrow\left\{{}\begin{matrix}a\cdot12=36\\b\cdot12=48\\c\cdot12=60\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}a=3\\b=4\\c=5\end{matrix}\right.\)

Ta thấy: \(\left\{{}\begin{matrix}x^2\ge0\\\left|y-2\right|\ge0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x^2\ge0\\3\left|y-2\right|\ge0\end{matrix}\right.\)

\(\Rightarrow x^2+3\left|y-2\right|\ge0\)

\(\Rightarrow x^2+3\left|y-2\right|-1\ge-1\)

Xảy ra khi \(\left\{{}\begin{matrix}x^2=0\\3\left|y-2\right|=0\end{matrix}\right.\)\(\Rightarrow\left\{{}\begin{matrix}x=0\\y=2\end{matrix}\right.\)