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\(Xét:\dfrac{\sqrt{x}}{\sqrt{x}+1}\) ta thấy rõ ràng : \(\sqrt{x}\ge0\)
\(\Rightarrow\sqrt{x}+1\ge1\)
\(\Rightarrow\sqrt{x}\) không thể : \(\ge\sqrt{x}+1\)
Do đó : \(0< \dfrac{\sqrt{x}}{\sqrt{x}+1}< 1\)
\(\dfrac{\sqrt{x}}{\sqrt{x}+1}\left(ĐK:x\ge0\right)\\ =\dfrac{\sqrt{x}+1}{\sqrt{x}+1}-\dfrac{1}{\sqrt{x}+1}\\ =1-\dfrac{1}{\sqrt{x}+1}\)
Ta thấy :
\(1>0,\sqrt{x}+1\ge1>0\forall x\ge0\\ =>\dfrac{1}{\sqrt{x}+1}>0\\ =>-\dfrac{1}{\sqrt{x}+1}< 0\\ =>1-\dfrac{1}{\sqrt{x}+1}< 1\\ =>\dfrac{\sqrt{x}}{\sqrt{x}+1}< 1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=x+\sqrt{x}=\sqrt{x}\left(\sqrt{x}+1\right).\)
Vì \(\sqrt{x}\ge0\)\(\Rightarrow B_{min}\)\(=0\Leftrightarrow\sqrt{x}\left(\sqrt{x}+1\right)=0\)
\(\Rightarrow\hept{\begin{cases}\sqrt{x}=0\\\sqrt{x}+1=0\end{cases}\Rightarrow\hept{\begin{cases}x=0\\x\in\varnothing\end{cases}}}\)
Vậy \(B_{min}=0\Leftrightarrow x=0\)
\(B=x+\sqrt{x}\)
\(B=\left(\sqrt{x}\right)^2+2\cdot\frac{1}{2}\sqrt{x}+\left(\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\)
\(B=\left(\sqrt{x}+\frac{1}{2}\right)^2-\left(\frac{1}{2}\right)^2\)
\(B=\left(\sqrt{x}+\frac{1}{2}\right)^2-\frac{1}{4}\)
Có \(\left(\sqrt{x}+\frac{1}{2}\right)^2\ge0\)
\(\Rightarrow\left(\sqrt{x}+\frac{1}{2}\right)^2-\frac{1}{4}\ge-\frac{1}{4}\)
\(\Rightarrow GTNN\left(\sqrt{x}+\frac{1}{2}\right)^2-\frac{1}{4}=-\frac{1}{4}\)
\(\Rightarrow GTNNx+\sqrt{x}=-\frac{1}{4}\)
với \(\left(\sqrt{x}+\frac{1}{2}\right)^2=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(C=x-2\sqrt{xy}+3y-2\sqrt{x}+1\)
\(=\left(\sqrt{x}-\sqrt{y}\right)^2+2y-2\left(\sqrt{x}-\sqrt{y}\right)-2\sqrt{y}+1\)
\(=\left(\sqrt{x}-\sqrt{y}\right)^2-2\left(\sqrt{x}-\sqrt{y}\right)+1+2\left(y-\sqrt{y}+\frac{1}{4}\right)-\frac{1}{2}\)
\(=\left(\sqrt{x}-\sqrt{y}-1\right)^2+2\left(\sqrt{y}-\frac{1}{2}\right)^2-\frac{1}{2}\ge\frac{-1}{2}\)
Đến đây dễ rồi
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: x>0
\(\dfrac{x+3}{\sqrt{x}}=\dfrac{x}{\sqrt{x}}+\dfrac{3}{\sqrt{x}}=\sqrt{x}+\dfrac{3}{\sqrt{x}}>=2\cdot\sqrt{\sqrt{x}\cdot\dfrac{3}{\sqrt{x}}}=2\sqrt{3}\)
Dấu '=' xảy ra khi \(\left(\sqrt{x}\right)^2=3\)
=>x=3
![](https://rs.olm.vn/images/avt/0.png?1311)