\(\frac{1}{1+a^2}+\frac{1}{1+b^2}\ge\frac{2}{1+ab}\)

\(\Leftrightarrow\frac{\left(1+b^2\right)\left(1+ab\right)+\left(1+a^2\right)\left(1+ab\right)-2\left(1+a^2\right)\left(1+b^2\right)}{\left(1+a^2\right)\left(1+b^2\right)\left(1+ab\right)}\ge0\)

\(\Leftrightarrow ab^3+b^2+ab+1+a^3b+a^2+ab+1-2a^2b^2-2a^2-2b^2-2\ge0\)

\(\Leftrightarrow a^3b+ab^3-2a^2b^2-a^2-b^2+2ab\ge0\)

\(\Leftrightarrow ab\left(a-b\right)^2-\left(a-b\right)^2\ge0\)

\(\Leftrightarrow\left(ab-1\right)\left(a-b\right)^2\ge0\)đúng do \(ab\ge1,\left(a-b\right)^2\ge0\).

Do biến đổi tương đương, bất đẳng thức cuối đúng nên bất đẳng thức cần chứng minh cũng đúng. 

Ta có đpcm. 

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

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

\(P=\frac{2\sqrt{x}+2y\sqrt{x}}{1-xy}.\frac{1-xy}{1+xy+x+y}\)

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

\(P=\frac{2\sqrt{x}}{x+1}\)

\(P=\frac{2}{\sqrt{x}+\frac{1}{\sqrt{x}}}\)

\(\sqrt{x}+\frac{1}{\sqrt{x}}\ge2\sqrt{\sqrt{x}.\frac{1}{\sqrt{x}}}=2\)

dấu "=" xảy ra khi x =1

\(P=\frac{2}{\sqrt{x}+\frac{1}{\sqrt{x}}}\le\frac{2}{2}=1\)

\(< =>MAX:P=1\)

Cảm ơn

\(=\sqrt{6-2\sqrt{30}+5}-\sqrt{5+2\sqrt{30}+6}\)

\(=\sqrt{\left(\sqrt{6}-\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{6}\right)^2}\)

\(=\sqrt{6}-\sqrt{5}-\sqrt{5}-\sqrt{6}\)

\(=-2\sqrt{5}\)

a) \(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)

\(=0.\sqrt{5}=0\)

b) \(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}=\frac{\left(5-2\sqrt{6}\right)\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right)\left(\sqrt{2}+\sqrt{3}\right)}=\frac{\left(5\sqrt{2}+5\sqrt{3}-4\sqrt{3}-6\sqrt{2}\right)}{-1}\)

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

a,\(\left(\sqrt{125}+\sqrt{45}-2\sqrt{80}\right).\sqrt{5}\)

\(=\left(5\sqrt{5}+3\sqrt{5}-8\sqrt{5}\right).\sqrt{5}\)

\(=0.\sqrt{5}\)

\(=0\)

b,\(\frac{5-2\sqrt{6}}{\sqrt{2}-\sqrt{3}}\)

\(=\frac{\left(5-2\sqrt{6}\right).\left(\sqrt{2}+\sqrt{3}\right)}{\left(\sqrt{2}-\sqrt{3}\right).\left(\sqrt{2}+\sqrt{3}\right)}\)

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

\(=\sqrt{2}-\sqrt{3}\)

\(A=\frac{\sqrt{x}-1}{\sqrt{x}}-9\sqrt{x}\)

\(A=1-\frac{1}{\sqrt{x}}-9\sqrt{x}\)

\(A=1-\left(\frac{1}{\sqrt{x}}+9\sqrt{x}\right)\)

\(\frac{1}{\sqrt{x}}+9\sqrt{x}\ge2\sqrt{\frac{1}{\sqrt{x}}.9\sqrt{x}}\)(cô - si)

\(2\sqrt{9}=6\)

\(< =>1-\left(\frac{1}{\sqrt{x}}+9\sqrt{x}\right)\le1-6=-5\)

dấu "=" xảy \(\frac{1}{\sqrt{x}}=9\sqrt{x}< =>x=\frac{1}{9}\)

\(< =>MAX:A=-5\)