a/ \(P=\frac{1}{\sqrt{xy}}\)

b/ \(x^3=8-6x\)

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

a/ Bạn tự tìm ĐKXĐ

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

Xét 

\(=\frac{\sqrt{x}-x\sqrt{y}+1-\sqrt{xy}+xy+\sqrt{xy}+x\sqrt{y}+\sqrt{x}+1-xy}{\left(1+\sqrt{xy}\right)\left(1-\sqrt{xy}\right)}\)

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

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

\(=\frac{xy-1-xy-\sqrt{xy}-x\sqrt{y}-\sqrt{x}-x\sqrt{y}+\sqrt{x}-\sqrt{xy}+1}{\left(\sqrt{xy}-1\right)\left(\sqrt{xy}+1\right)}\)

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

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

b/ Áp dụng BĐT \(\left(a+b\right)^2\ge4ab\) với \(a=\frac{1}{\sqrt{x}},b=\frac{1}{\sqrt{y}}\) được : 

\(A=\frac{1}{\sqrt{x}.\sqrt{y}}\le\frac{1}{4}\left(\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}\right)^2=\frac{1}{4}.6^2=9\)

Dấu "=" xảy ra khi \(\hept{\begin{cases}\sqrt{x}=\sqrt{y}\\\frac{1}{\sqrt{x}}+\frac{1}{\sqrt{y}}=6\end{cases}}\Leftrightarrow x=y=\frac{1}{9}\)

Vậy ........................................................

\(D=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}-\sqrt{xy}+1}{\sqrt{xy}-1}\right)\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}-\frac{\sqrt{xy}+\sqrt{x}-\sqrt{xy}+1}{\sqrt{xy}-1}\right)\)

\(D=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{x}+1}{\sqrt{xy}-1}\right)\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}-\frac{\sqrt{x}+1}{\sqrt{xy}-1}\right)\)

\(D=\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{xy}+1\right)^2}-\frac{\left(\sqrt{x}+1\right)^2}{\left(\sqrt{xy}-1\right)^2}\)

\(D=\left(\sqrt{x}+1\right)^2\left(\frac{1}{\left(\sqrt{xy}+1\right)^2}-\frac{1}{\left(\sqrt{xy}-1\right)^2}\right)\)

\(D=\left(\sqrt{x}+1\right)^2\cdot\frac{xy+1-2\sqrt{xy}-xy-1-2\sqrt{xy}}{\left(xy-1\right)^2}\)

\(D=\frac{\left(\sqrt{x}+1\right)^2\cdot\left(-4\sqrt{xy}\right)}{\left(xy-1\right)^2}\)

\(P=\left(\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\right):\left(1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\right)\)

+) Đặt \(Q=\frac{\sqrt{x}+1}{\sqrt{xy}+1}+\frac{\sqrt{xy}+\sqrt{x}}{1-\sqrt{xy}}+1\)

\(Q=\frac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{xy-1}-\frac{\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)}{xy-1}+\frac{xy-1}{xy-1}\)

\(Q=\frac{x\sqrt{y}-\sqrt{x}+\sqrt{xy}-1-xy-x\sqrt{y}-\sqrt{xy}-\sqrt{x}+xy-1}{xy-1}\)

\(Q=\frac{-2-2\sqrt{x}}{xy-1}\)

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

+) Đặt \(K=1-\frac{\sqrt{xy}+\sqrt{x}}{\sqrt{xy}-1}-\frac{\sqrt{x}+1}{\sqrt{xy}+1}\)

\(K=\frac{xy-1}{xy-1}-\frac{\left(\sqrt{xy}+\sqrt{x}\right)\left(\sqrt{xy}+1\right)}{xy-1}-\frac{\left(\sqrt{x}+1\right)\left(\sqrt{xy}-1\right)}{xy-1}\)

\(K=\frac{xy-1-xy-x\sqrt{y}-\sqrt{xy}-\sqrt{x}-x\sqrt{y}+\sqrt{x}-\sqrt{xy}+1}{xy-1}\)

\(K=\frac{-2x\sqrt{y}-2\sqrt{xy}}{xy-1}\)

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

Ta có : \(P=Q:K\)

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

\(\Leftrightarrow P=\frac{-2\left(\sqrt{x}+1\right)\left(xy-1\right)}{-2\sqrt{xy}\left(\sqrt{x}+1\right)\left(xy-1\right)}\)

\(\Leftrightarrow P=\frac{1}{\sqrt{xy}}\)

Vậy...

Trần Thanh Phương

b, Tính P khi \(x=\sqrt[3]{4+2\sqrt{6}}+\sqrt[3]{4-2\sqrt{6}};y=x^2+6\)