Với x > = 0 

a, Ta có : \(\sqrt{x}+3\ge3\Rightarrow P=\frac{3}{\sqrt{x}+3}\le\frac{3}{3}=1\)

Dấu ''='' xảy ra khi x = 0 

Vậy GTLN của P bằng 1 tại x = 0 

6 nhé 

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k cho mk 

HT

có phải = 1447728866271753e+ 16 không

điều kiện : \(x\ge1\)

ta có : \(x+6=6\sqrt{x-1}\Leftrightarrow\left(x+6\right)^2=36\left(x-1\right)\Leftrightarrow x^2-24x+72=0\Leftrightarrow x=12\pm6\sqrt{2}\)

cả hai nghiệm đều thỏa mãn

ta có :

\(P=\frac{\sqrt{x}-1}{\sqrt{x}+2}< \frac{1}{2}\Leftrightarrow2\sqrt{x}-2< \sqrt{x}+2\Leftrightarrow\sqrt{x}< 4\Leftrightarrow0\le x< 16\)

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. \(A\text{ nguyên }\Rightarrow\frac{1}{\sqrt{x}}\text{ nguyên }\Rightarrow x=1\)

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 \(6-4\sqrt{2}+\frac{3}{2}\le S\le6+4\sqrt{2}+\frac{3}{2}\)