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\(A\le\sqrt{\left(3^2+4^2\right)\left(x-1\right)\left(5-x\right)}=10\)
\(A_{max}=10\) khi \(\dfrac{\sqrt{x-1}}{3}=\dfrac{\sqrt{5-x}}{4}\Rightarrow x=\dfrac{61}{25}\)
\(A=3\left(\sqrt{x-1}+\sqrt{5-x}\right)+\sqrt{5-x}\ge3\left(\sqrt{x-1}+\sqrt{5-x}\right)\ge3\sqrt{x-1+5-x}=6\)
\(A_{min}=6\) khi \(x=5\)
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Câu 2:
\(C=-x+\sqrt{x}\)
\(=-\left(x-\sqrt{x}+\dfrac{1}{4}\right)+\dfrac{1}{4}\)
\(=-\left(\sqrt{x}-\dfrac{1}{2}\right)^2+\dfrac{1}{4}\le\dfrac{1}{4}\forall x\)
Dấu '=' xảy ra khi \(x=\dfrac{1}{4}\)
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để biểu thức C xác định thì xảy ra đồng thời
- x-2>=0
- 5-x>=0
=>2=<x=<5
thay x=2;3;4;5
tim ra gia tri nho nhat va lon nhat
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a . ta có : \(1\le1+\sqrt{2-x}\Rightarrow GTNN=1\)
\(-2\le\sqrt{x-3}-2\Rightarrow GTNN=-2\)
b. \(0\le\sqrt{4-x^2}\le2\)
\(\sqrt{2x^2-x+3}=\sqrt{2\left(x^2-\frac{x}{2}+\frac{1}{16}\right)+\frac{23}{8}}=\sqrt{2\left(x-\frac{1}{4}\right)^2+\frac{23}{8}}\ge\frac{\sqrt{46}}{4}\)
vậy \(GTNN=\frac{\sqrt{46}}{4}\)
ta có : \(0\le-x^2+2x+5=-\left(x-1\right)^2+6\le6\)
\(\Rightarrow1-\sqrt{6}\le1-\sqrt{-x^2+2x+5}\le1\)Vậy \(\hept{\begin{cases}GTNN=1-\sqrt{6}\\GTLN=1\end{cases}}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(B=\dfrac{x-\sqrt[]{x}}{\sqrt[]{x}-\left(x+1\right)}\)
\(B\) xác định \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt[]{x}-\left(x+1\right)\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x^2+x+1\ne0,\forall x\in R\end{matrix}\right.\) \(\Leftrightarrow x\ge0\)
\(\Leftrightarrow B=\dfrac{x-\sqrt[]{x}+1-1}{-\left(x-\sqrt[]{x}+1\right)}\)
\(\Leftrightarrow B=-1+\dfrac{1}{x-\sqrt[]{x}+1}\)
\(\Leftrightarrow B=-1+\dfrac{1}{x-\sqrt[]{x}+\dfrac{1}{4}-\dfrac{1}{4}+1}\)
\(\Leftrightarrow B=-1+\dfrac{1}{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\)
mà \(\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}\ge\dfrac{3}{4},\forall x\ge0\)
\(\Rightarrow B=-1+\dfrac{1}{\left(\sqrt[]{x}-\dfrac{1}{2}\right)^2+\dfrac{3}{4}}\le-1+\dfrac{4}{3}=\dfrac{1}{3}\)
\(\Rightarrow GTLN\left(B\right)=\dfrac{1}{3}\left(tại.x=\dfrac{1}{4}\right)\)