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a) ĐKXĐ: \(x>0\)
\(A=\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\)
\(=x+\sqrt{x}-2\sqrt{x}-1+1=x-\sqrt{x}\)
\(A=x-\sqrt{x}=2\)
\(\Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\)
\(\Leftrightarrow\sqrt{x}=2\Leftrightarrow x=4\left(tm\right)\)(do \(\sqrt{x}+1\ge1>0\))
b) \(A=x-\sqrt{x}=\sqrt{x}\left(\sqrt{x}-1\right)>0\)(do \(x>1\))
\(\Leftrightarrow A=x-\sqrt{x}=\left|A\right|\)
c) \(A=x-\sqrt{x}=\left(x-\sqrt{x}+\dfrac{1}{4}\right)-\dfrac{1}{4}\)
\(=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\)
\(minA=-\dfrac{1}{4}\Leftrightarrow\sqrt[]{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
\(a,A=\dfrac{x\left(x\sqrt{x}+1\right)}{x-\sqrt{x}+1}-\dfrac{\sqrt{x}\left(2\sqrt{x}+1\right)}{\sqrt{x}}+1\left(x>0\right)\\ A=\dfrac{x\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{x-\sqrt{x}+1}-2\sqrt{x}-1+1\\ A=x+\sqrt{x}-2\sqrt{x}=x-\sqrt{x}\\ A=2\Leftrightarrow x-\sqrt{x}-2=0\\ \Leftrightarrow\left(\sqrt{x}-2\right)\left(\sqrt{x}+1\right)=0\\ \Leftrightarrow\sqrt{x}=2\left(\sqrt{x}>0\right)\\ \Leftrightarrow x=4\left(tm\right)\)
\(b,x>1\Leftrightarrow\sqrt{x}-1>0\\ \Leftrightarrow\left|A\right|=\left|x-\sqrt{x}\right|=\left|\sqrt{x}\left(\sqrt{x}-1\right)\right|=\sqrt{x}\left(\sqrt{x}-1\right)=A\left(\sqrt{x}>0\right)\)
\(c,A=x-\sqrt{x}+\dfrac{1}{4}-\dfrac{1}{4}=\left(\sqrt{x}-\dfrac{1}{2}\right)^2-\dfrac{1}{4}\ge-\dfrac{1}{4}\\ A_{min}=-\dfrac{1}{4}\Leftrightarrow\sqrt{x}=\dfrac{1}{2}\Leftrightarrow x=\dfrac{1}{4}\left(tm\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
b: Thay \(x=7-2\sqrt{6}\) vào A, ta được:
\(A=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-7+2\sqrt{6}-5\left(\sqrt{6}+1\right)-1}\)
\(=\dfrac{3\cdot\left(\sqrt{6}-1\right)}{-8+2\sqrt{6}-5\sqrt{6}-5}\)
\(=\dfrac{-3\sqrt{6}+3}{13+3\sqrt{6}}=\dfrac{93-48\sqrt{6}}{115}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A^2=\left(\sqrt{1-x}+\sqrt{1+x}\right)^2\le\left(1^2+1^2\right)\left(1-x+1+x\right)=4\\ \Leftrightarrow A\le2\\ A_{max}=2\Leftrightarrow1-x=1+x\Leftrightarrow x=0\\ A^2=2+2\sqrt{1-x^2}\ge2\\ \Leftrightarrow A\ge\sqrt{2}\\ A_{min}=\sqrt{2}\Leftrightarrow1-x^2=0\Leftrightarrow\left[{}\begin{matrix}x=1\\x=-1\end{matrix}\right.\)
Vậy \(\sqrt{2}\le A\le2\)
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A = \(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\) ; Đk \(x\) \(\ge\) 0
A = 2 - \(\dfrac{3}{\sqrt{x}+1}\)
\(\sqrt{x}\) \(\ge\) 0 => \(\sqrt{x}\) + 1 \(\ge\) 1
\(\dfrac{3}{\sqrt{x}+1}\) \(\le\) \(\dfrac{3}{1}\)
\(\dfrac{3}{\sqrt{x}+1}\) \(\le\) 3
- \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) -3
2 - \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) 2 - 3
2 - \(\dfrac{3}{\sqrt{x}+1}\) \(\ge\) -1
A(min ) = -1 dấu bằng xảy ra khi \(\sqrt{x}\) = 0 hay \(x\) = 0
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\sqrt{A}\ge0\) ; \(\forall A\) nên GTNN của \(\sqrt{A}\) là \(0\)
Dấu "=" xảy ra khi \(x=0\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Câu 1:
\(a-\sqrt{a}+1=a-2.\sqrt{a}.\frac{1}{2}+\frac{1}{2^2}+\frac{3}{4}\)
\(=(\sqrt{a}-\frac{1}{2})^2+\frac{3}{4}\)
Ta thấy \((\sqrt{a}-\frac{1}{2})^2\geq 0, \forall a\) không âm
\(\Rightarrow a-\sqrt{a}+1=(\sqrt{a}-\frac{1}{2})^2+\frac{3}{4}\geq \frac{3}{4}\)
Vậy GTNN của biểu thức là $\frac{3}{4}$. Dấu "=" xảy ra khi \((\sqrt{a}-\frac{1}{2})^2=0\Leftrightarrow a=\frac{1}{4}\)
Câu 2:
\(\sqrt{1+2a-a^2}=\sqrt{2-(a^2-2a+1)}=\sqrt{2-(a-1)^2}\)
Ta thấy \((a-1)^2\geq 0, \forall a\) thuộc tập xác định
\(\Rightarrow 2-(a-1)^2\leq 2\)
\(\Rightarrow \sqrt{1+2a-a^2}=\sqrt{2-(a-1)^2}\leq \sqrt{2}\)
Vậy GTLN của biểu thức là $\sqrt{2}$ khi \((a-1)^2=0\Leftrightarrow a=1\)
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\(A=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}+\frac{3\sqrt{x}+1}{1-x}\)
a) ĐKXĐ : \(\hept{\begin{cases}x\ge0\\x\ne1\end{cases}}\)
\(=\frac{\sqrt{x}+1}{\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+1}-\frac{3\sqrt{x}+1}{x-1}\)
\(=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{\left(\sqrt{x}-1\right)^2}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}+\frac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}-\frac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(=\frac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\frac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b) Với x = 4 thỏa mãn ĐKXĐ
\(A=\frac{2\sqrt{4}-1}{\sqrt{4}+1}=\frac{4-1}{2+1}=\frac{3}{3}=1\)
c) Chưa nghĩ ra :<
![](https://rs.olm.vn/images/avt/0.png?1311)
\(A\le\sqrt{\left(3^2+4^2\right)\left(x-1\right)\left(5-x\right)}=10\)
\(A_{max}=10\) 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 \(x=5\)
(x+√x+1/4)+3/4
(√x+1/2)2+3/4
ta có (√x+1/2)2>=0
(√x+1/2)2+3/4>=3/4
minA=3/4 và xảy ra khi √x+1/2=0=>√x=-1/2=>x=1/4