\(A=\frac{x^2+2x+3}{x^2+4x+4}-\frac{2}{3}+\frac{2}{3}\)

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

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

\(\hept{\begin{cases}\left(x-1\right)^2\ge0\\\left(x+2\right)^2\ge0\end{cases}\Rightarrow\frac{\left(x-1\right)^2}{\left(x+2\right)^2}\ge0}\)

Dấu '' ='' xảy ra khi và chỉ khi  x=1

=> Min A =2/3 khi x=1

ĐK: \(x\ge0;x\ne1\)

Ta có: \(P=\text{[}\frac{\sqrt{x}-2}{x-1}-\frac{\left(\sqrt{x}+2\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}+1\right)^2\left(\sqrt{x}-1\right)}\text{]}\left(\frac{1-x}{\sqrt{2}}\right)^2\)

\(=\text{[}\frac{\sqrt{x}-2}{x-1}-\frac{x+\sqrt{x}-2}{\left(x-1\right)\left(\sqrt{x}+1\right)}\text{]}\frac{\left(x-1\right)^2}{2}\)

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

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

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

Ta có \(x^2-x-1=0\Rightarrow x^2-x=1\Rightarrow\left(x^2-x\right)^3=1\)

\(\Rightarrow x^6-3x^5+3x^4-x^3=1\)

Mặt khác \(x^2-x-1-0\Rightarrow x^2=x+1\)

\(\Rightarrow x^6=\left(x+1\right)^3=x^3+2=3x^2+3x+1\)

\(\Rightarrow P=\frac{1+2017}{1+2017}=1\)

\(cos^4x-sin^4x=2cos^2x-1\) ( 1 ) 

\(\left(cos^2x\right)^2-\left(sin^2x\right)^2=2cos^2x-1\) 

\(\left(cos^2x-sin^2x\right)\left(cosx^2+sin^2x\right)=2cos^2x-1\)  

\(\left(cos^2x-sin^2x\right)\cdot1=2cos^2x-1\)

 \(cos^2x-sin^2x=2cos^2x-1\) 

\(cosx^2-\left(1-cos^2x\right)=2cos^2x-1\)  

\(cos^2x-1+cos^2x=2cos^2x-1\)  

\(2cos^2x-1=2cos^2x-1\) 

\(0=0\left(llđ\right)\) ( tới hàng trên luôn luôn đúng cũng được chứ không cần tới hàng này nha ) 

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