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Đề là: \(A=\sqrt{x-2\sqrt{x-3}}\) đúng ko em?
ĐKXĐ: \(x\ge3\)
\(A=\sqrt{x-3-2\sqrt{x-3}+1+2}=\sqrt{\left(\sqrt{x-3}-1\right)^2+2}\ge\sqrt{2}\)
\(A_{min}=\sqrt{2}\) khi \(x=4\)
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A = \(\frac{\sqrt{x}}{\sqrt{x}+3}=\frac{\sqrt{x}-3+3}{\sqrt{x}+3}=1-\frac{3}{\sqrt{x}+3}\)
để A min => \(\frac{3}{\sqrt{x}+3}\) max => \(\sqrt{x}+3\) min
thấy \(\sqrt{x}\ge0\Rightarrow\sqrt{x}+3\ge3\)
\(\Rightarrow\frac{3}{\sqrt{x}+3}\le1\Rightarrow1-\frac{3}{\sqrt{x}+3}\ge0\)
vậy min A = 0 khi x = 0
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\(y=\sqrt{x-1}+\sqrt{9-x}\)(đk: \(9\ge x\ge1\))
=> \(y\ge\sqrt{x-1+9-x}=\sqrt{8}\)
Dấu "=" xảy ra khi x =1 hoặc x= 9
Vậy y min = \(\sqrt{8}\)khi x =1 hoặc x = 9
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\(B=x+\sqrt{x}=\sqrt{x}\left(\sqrt{x}+1\right).\)
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 \(\left(\sqrt{x}+\frac{1}{2}\right)^2=0\)
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