\(P=\frac{\sqrt{a+1}+\sqrt{a-1}}{\sqrt{a+1}-\sqrt{a-1}}\left(a-\sqrt{\left(a-1\right)\left(a+1\right)}\right)\)

\(P=\frac{\left(\sqrt{a+1}+\sqrt{a-1}\right)^2}{a+1-a+1}\left(a-\sqrt{\left(a-1\right)\left(a+1\right)}\right)\)

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

\(P=\left(a+\sqrt{a^2-1}\right)\left(a-\sqrt{a^2-1}\right)=a^2-a^2+1=1\)

Vô Pitago mà hỏi

là sao?

Ta có:

\(\left(x+y\right)=\left(x+y+z\right)^2\left(x+y\right)\)

\(\ge4\left(x+y\right)^2z\ge16xyz\)

Dấu = xảy ra khi \(\hept{\begin{cases}x=y=\frac{1}{4}\\z=\frac{1}{2}\end{cases}}\)

là sao alibaba nguyễn mk ko hiểu

và tìm điều kiện 

 đk \(x\ge0\)

\(\frac{\sqrt{3x}-3}{3+\sqrt{3x}}=-\frac{1}{5}\)

\(\Leftrightarrow\frac{\left(\sqrt{3x}-3\right)^2}{\left(\sqrt{3x}-3\right)\left(3+\sqrt{3x}\right)}=-\frac{1}{5}\)

\(\Leftrightarrow\frac{3x-6\sqrt{3x}+9}{3x-9}=-\frac{1}{5}\)

\(\Leftrightarrow\frac{\left(x-2\sqrt{3x}+3\right)}{x-3}=-\frac{1}{5}\)

\(\Leftrightarrow5\left(x-2\sqrt{3x}+3\right)=3-x\)

\(\Leftrightarrow5x-10\sqrt{3x}+15=3-x\)

\(\Leftrightarrow6x-2.5\sqrt{3x}+12=0\)

\(tan^2x-sin^2x=\frac{sin^2x}{cos^2x}-sin^2x\)

\(=sin^2x.\left(\frac{1}{cos^2x}-1\right)=sin^2x.\frac{sin^2x}{cos^2x}=tan^2x.sin^2x\)

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

=> \(A=\left(2\sqrt{xy}+x+y\right).\frac{1}{\left(\sqrt{x}+\sqrt{y}\right)^2}\)

=> \(A=\frac{\left(\sqrt{x}+\sqrt{y}\right)^2}{\left(\sqrt{x}+\sqrt{y}\right)^2}=1\)

ĐS: A=1