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Lời giải:
ĐKXĐ: $-2\leq x\leq 38$
Áp dụng BĐT Bunhiacopxky:
$A^2=(\sqrt{x+2}+\sqrt{38-x})^2\leq (x+2+38-x)(1+1)=80$
$\Rightarrow A\leq \sqrt{80}$
Vậy $A_{\max}=\sqrt{80}$
![](https://rs.olm.vn/images/avt/0.png?1311)
DKXD của A, ta có \(x^{2\le5\Rightarrow-\sqrt{5}\le x\le\sqrt{5}}\)
mà \(3x\ge-3\sqrt{5}\)
mặt kkhác \(\sqrt{5-x^2}\ge0\Rightarrow A=3x+x\sqrt{5-x^2}\ge-3\sqrt{5}\)
min A= \(-3\sqrt{5}\)\(\Leftrightarrow x=-\sqrt{5}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(\forall x\in R\Rightarrow A=\dfrac{\sqrt{x}}{x-2\sqrt{x}+9}\Leftrightarrow A\left(x-2\sqrt{x}+9\right)=\sqrt{x}\)
\(\Leftrightarrow Ax-2A\sqrt{x}-\sqrt{x}+9A=0\)
\(\Leftrightarrow A\sqrt{x}^2-\sqrt{x}\left(2A+1\right)+9A=0\)
\(\Rightarrow\Delta\ge0\Rightarrow\left(2A+1\right)^2-36A^2=-32A^2+4A+1\ge0\Rightarrow-\dfrac{1}{8}\le A\le\dfrac{1}{4}\Rightarrow A\le\dfrac{1}{4}\Rightarrow MaxA=\dfrac{1}{4}\)
\(dấu"="\) \(xảy\) \(ra\Leftrightarrow x=9\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(Đk:x\ge2\)
Đặt \(A=\sqrt{x-2}+2\sqrt{x+1}+2019-x\)
\(2A=2\sqrt{x-2}+4\sqrt{x+1}+4038-2x\)
\(=\left[\left(-x+2\right)+2\sqrt{x-2}-1\right]+\left[\left(-x-1\right)+4\sqrt{x+1}-4\right]+4042\)
\(=-\left[\left(x-2\right)-2\sqrt{x-2}+1\right]-\left[\left(x+1\right)-4\sqrt{x+1}+4\right]+4042\)
\(=-\left(\sqrt{x-2}-1\right)^2-\left(\sqrt{x+1}-2\right)^2+4042\le4042\)
\(\Rightarrow A\le2021\)
Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}\sqrt{x-2}-1=0\\\sqrt{x+1}-2=0\end{matrix}\right.\Leftrightarrow x=3\)
Vậy \(Max\) của biểu thức trên là 2021, đạt tại x=3.
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x\ne9\end{matrix}\right.\)
\(P=\dfrac{\sqrt{x}}{\sqrt{x}-3}\)
\(=\dfrac{\sqrt{x}-3+3}{\sqrt{x}-3}\)
\(=1+\dfrac{3}{\sqrt{x}-3}\)
\(\sqrt{x}-3>=-3\forall x\) thỏa mãn ĐKXĐ
=>\(\dfrac{3}{\sqrt{x}-3}< =\dfrac{3}{-3}=-1\forall x\) thỏa mãn ĐKXĐ
=>\(\dfrac{3}{\sqrt{x}-3}+1< =-1+1=0\forall x\) thỏa mãn ĐKXĐ
=>P<=0 với mọi x thỏa mãn ĐKXĐ
Dấu '=' xảy ra khi x=0
![](https://rs.olm.vn/images/avt/0.png?1311)
Lời giải:
Đặt $\sqrt{2+x}=a; \sqrt{2-x}=b$. ĐK: $a,b\geq 0$
$a^2+b^2=4$
Gọi biểu thức cần tìm min max là $D$
$D=a+b-ab=(a-2)(2-b)+4-(a+b)$
Vì $a^2+b^2=4\Rightarrow a,b\leq 2$
$\Rightarrow (a-2)(2-b)\leq 0$
Mặt khác: $a^2+b^2=4\Rightarrow (a+b)^2=4+2ab\geq 4$
$\Rightarrow a+b\geq 2$
Do đó: $D=(a-2)(2-b)+4-(a+b)\leq 4-(a+b)\leq 2$
Vậy $D_{\max}=2$ khi $x=\pm 2$
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$4=a^2+b^2\geq 2ab\Rightarrow ab\leq 2$
$D=a+b-ab=\sqrt{4+2ab}-ab$
$=\sqrt{4+2ab}-2\sqrt{2}-(ab-2)+2\sqrt{2}-2$
$=\frac{2(ab-2)}{\sqrt{4+2ab}+2\sqrt{2}}-(ab-2)+2\sqrt{2}-2$
$=(ab-2)(\frac{2}{\sqrt{4+2ab}+2\sqrt{2}}-1)+2\sqrt{2}-2$
Vì $ab\leq 2\rightarrow ab-2\leq 0$
$ab\geq 0\Rightarrow \frac{2}{\sqrt{4+2ab}+2\sqrt{2}}-1 <\frac{2}{\sqrt{4}+2\sqrt{2}}-1<0$
$\Rightarrow D\geq 0+2\sqrt{2}-2=2\sqrt{2}-2$
Vậy $D_{\min}=2\sqrt{2}-2$ khi $x=0$
![](https://rs.olm.vn/images/avt/0.png?1311)
\(D=\dfrac{\sqrt{x}-x-7}{\sqrt{x}+1}=\dfrac{\left(\sqrt{x}+1\right)-\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)-9}{\sqrt{x}+1}=1-\sqrt{x}+1-\dfrac{9}{\sqrt{x}+1}\)
\(=3-\left[\left(\sqrt{x}+1\right)+\dfrac{9}{\sqrt{x}+1}\right]\)\(\le3-2\sqrt{\left(\sqrt{x}+1\right).\dfrac{9}{\sqrt{x}+1}}\) ( BĐT AM-GM)
\(\Leftrightarrow D\le-3\)
Dấu "=" xảy ra khi \(\sqrt{x}+1=\dfrac{9}{\sqrt{x}+1}\Leftrightarrow x=4\)
Vậy \(max_D=-3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ĐKXĐ: x>=0
\(M=\dfrac{\sqrt{x}+2+1}{\sqrt{x}+2}=1+\dfrac{1}{\sqrt{x}+2}\)
\(\sqrt{x}+2>=2\)
=>\(\dfrac{1}{\sqrt{x}+2}< =\dfrac{1}{2}\)
=>\(M=1+\dfrac{1}{\sqrt{x}+2}< =\dfrac{1}{2}+1=\dfrac{3}{2}\)
Dấu = xảy ra khi x=0