làm phần E g thôi nhé
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![](https://rs.olm.vn/images/avt/0.png?1311)
điều kiện :\(\hept{\begin{cases}x-\sqrt{x}\ne0\\x\ge0\\\sqrt{x}+1\ne0\end{cases}}\Leftrightarrow0< x\ne1\)
\(A=\left(\frac{1+\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}\right):\frac{\sqrt{x}+1}{\left(\sqrt{x}-1\right)^2}=\frac{\sqrt{x}-1}{\sqrt{x}}=1-\frac{1}{\sqrt{x}}\)
b. \(A=\frac{1}{3}=1-\frac{1}{\sqrt{x}}\Leftrightarrow\frac{1}{\sqrt{x}}=\frac{2}{3}\Leftrightarrow x=\frac{9}{4}\)
c. \(x=\sqrt{4+3+4\sqrt{3}}+\sqrt{4+3-4\sqrt{3}}=2+\sqrt{3}+2-\sqrt{3}=4\Rightarrow A=1-\frac{1}{\sqrt{4}}=\frac{1}{2}\)
d.\(A=-1=1-\frac{1}{\sqrt{x}}\Leftrightarrow\frac{1}{\sqrt{x}}=2\Leftrightarrow x=\frac{1}{4}\)
e. \(A>\frac{1}{2}\Leftrightarrow\frac{1}{\sqrt{x}}< \frac{1}{2}\Leftrightarrow\sqrt{x}>2\Leftrightarrow x>4\)
e. \(A\text{ nguyên }\Rightarrow\frac{1}{\sqrt{x}}\text{ nguyên }\Rightarrow x=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có :
\(S=4-2\left(x+y\right)+xy=4-2\left(x+y\right)+xy+\frac{x^2+y^2-1}{2}\)
\(=\frac{\left(x+y\right)^2}{2}-2\left(x+y\right)+\frac{7}{2}=\frac{1}{2}\left(x+y-2\right)^2+\frac{3}{2}\)
mà ta có : \(\left(x+y\right)^2\le2\left(x^2+y^2\right)=2\Rightarrow-\sqrt{2}\le x+y\le\sqrt{2}\)
\(\Rightarrow-2-\sqrt{2}\le x+y-2\le\sqrt{2}-2< 0\)
\(\Rightarrow\left(2-\sqrt{2}\right)^2\le\left(x+y-2\right)^2\le\left(\sqrt{2}+2\right)^2\)
hay \(6-4\sqrt{2}+\frac{3}{2}\le S\le6+4\sqrt{2}+\frac{3}{2}\)
![](https://rs.olm.vn/images/avt/0.png?1311)
TA CÓ \(2\sqrt{33}>2\sqrt{25}=2\times5=10\)
---> \(2\sqrt{33}>10\)
XIN TIICK
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có :
\(5-\sqrt{8}< 5-\sqrt{4}=5-2=3\)
Vậy \(5-\sqrt{8}< 3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : 3 < 5 => \(\sqrt{3}< \sqrt{5}\)
=> \(-\sqrt{3}>-\sqrt{5}\)
TA CÓ \(3< 5\)
\(\Rightarrow\sqrt{3}< \sqrt{5}\)
\(\Rightarrow-\sqrt{3}>-\sqrt{5}\)
XIN TIICK
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
![](https://rs.olm.vn/images/avt/0.png?1311)
ta có :
\(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}< \frac{1}{2}\Leftrightarrow2\sqrt{x}-2< \sqrt{x}+2\Leftrightarrow\sqrt{x}< 4\Leftrightarrow0\le x< 16\)