i am in class 6

a/ \(P=\frac{1}{\sqrt{xy}}\)

b/ \(x^3=8-6x\)

\(\Rightarrow P=\frac{1}{\sqrt{x\left(x^2+6\right)}}=\frac{1}{\sqrt{x^3+6x}}=\frac{1}{\sqrt{8-6x+6x}}=\frac{1}{2\sqrt{2}}\)

1.

a,

\(A\text{ xác định }\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\x-\sqrt{x}\ne0\end{matrix}\right.\)

\(\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\\x\ne0\end{matrix}\right.\)

\(\text{Vậy A xác định }\Leftrightarrow x>0\text{ và }x\ne1\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)

\(=\left(\frac{\sqrt{x}.\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{2}{\sqrt{x}\left(\sqrt{x}-1\right)}\right).\left(\sqrt{x}-1\right)\)

\(=\frac{x-2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)=\frac{x-2}{\sqrt{x}}\)

b, \(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)

\(\Rightarrow\sqrt{x}=\sqrt{\left(\sqrt{2}-1\right)^2}=\left|\sqrt{2}-1\right|=\sqrt{2}-1\)

\(A=\frac{x-2}{\sqrt{x}}=\frac{3-2\sqrt{2}-2}{\sqrt{2}-1}\)

\(=\frac{1-2\sqrt{2}}{\sqrt{2}-1}=-\frac{\left(\sqrt{2}-1\right)\left(2+\sqrt{2}+1\right)}{\sqrt{2}-1}=-3-\sqrt{2}\)

1.a)\(ĐKXĐ:\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\x-\sqrt{x}\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\\x\ne0;x\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>0\\x\ne1\end{matrix}\right.\)

\(A=\left(\frac{\sqrt{x}}{\sqrt{x}-1}+\frac{2}{x-\sqrt{x}}\right):\frac{1}{\sqrt{x}-1}\)

\(=\frac{x+2}{\sqrt{x}\left(\sqrt{x}-1\right)}.\left(\sqrt{x}-1\right)\)

\(=\frac{x+2}{\sqrt{x}}\)

b)\(x=3-2\sqrt{2}=2-2\sqrt{2}+1=\left(\sqrt{2}-1\right)^2\)

Thay x vào A ta được

\(A=\frac{3-2\sqrt{2}+2}{\sqrt{\left(\sqrt{2}-1\right)^2}}\)

\(=\frac{\sqrt{2}+6-1-3\sqrt{2}}{\sqrt{2}-1}\)(vì \(\sqrt{2}>1\))

\(=\frac{\sqrt{2}\left(1+3\sqrt{2}\right)-\left(1+3\sqrt{2}\right)}{\sqrt{2}-1}\)

\(=\frac{\left(\sqrt{2}-1\right)\left(1+3\sqrt{2}\right)}{\sqrt{2}-1}=1+3\sqrt{2}\)

c)\(A=\frac{x+2}{\sqrt{x}}=\sqrt{x}+\frac{2}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{2}{\sqrt{x}}}=2\sqrt{2}\)

Dấu "=" xảy ra khi \(x=2\)

\(\Rightarrow Min_A=2\sqrt{2}\) khi x=2

ĐKXĐ : \(x,y>0\)

a/ \(A=\left(\sqrt{x}+\frac{y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\frac{x}{\sqrt{xy}+y}+\frac{y}{\sqrt{xy}-x}+\frac{x+y}{\sqrt{xy}}\right)\)

\(=\left(\frac{x+\sqrt{xy}+y-\sqrt{xy}}{\sqrt{x}+\sqrt{y}}\right):\left(\frac{x\sqrt{x}\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right).\sqrt{x}}-\frac{y\sqrt{y}\left(\sqrt{x}+\sqrt{y}\right)}{\sqrt{x}.\sqrt{y}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}-\frac{\left(x+y\right)\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{\sqrt{xy}\left(\sqrt{x}-\sqrt{y}\right)\left(\sqrt{x}+\sqrt{y}\right)}\right)\)

\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{x^2-x\sqrt{xy}-y\sqrt{xy}-y^2-x^2+y^2}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}=\frac{x+y}{\sqrt{x}+\sqrt{y}}:\frac{-\sqrt{xy}\left(x+y\right)}{\sqrt{xy}\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}\)

\(=\frac{x+y}{\sqrt{x}+\sqrt{y}}.\frac{-\left(\sqrt{x}+\sqrt{y}\right)\left(\sqrt{x}-\sqrt{y}\right)}{x+y}=\sqrt{y}-\sqrt{x}\)

b/ Ta có ; \(4+2\sqrt{3}=\left(\sqrt{3}+1\right)^2\)

\(\Rightarrow B=\sqrt{\left(\sqrt{3}+1\right)^2}-\sqrt{3}=\sqrt{3}+1-\sqrt{3}=1\)