\(\frac{1}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{4}\)

.......

..........

\(\frac{1}{\sqrt{5}+1}=\frac{\sqrt{5}-1}{\left(\sqrt{5}-1\right)\left(\sqrt{5}+1\right)}=\frac{\sqrt{5}-1}{5-1}=\frac{\sqrt{5}-1}{4}\)

\(A=\left(\frac{x+1}{x-1}-\frac{x-1}{x+1}+\frac{x^2-4x-1}{x^2-1}\right).\frac{x+2003}{x}\)ĐK : \(x\ne0;\pm1\)

\(=\left(\frac{x^2+2x+1-x^2+2x-1+x^2-4x-1}{\left(x-1\right)\left(x+1\right)}\right).\frac{2003}{x}\)

\(=\frac{x^2-1}{\left(x-1\right)\left(x+1\right)}.\frac{2003}{x}=\frac{2003}{x}\)

Ta có: \(M=\frac{a^2b+b^2a+1}{a+b}=\frac{ab\left(a+b\right)+1}{a+b}=ab+\frac{1}{a+b}=3+\frac{1}{a+b}\)(vì ab = 3)

Lại có: \(a+b\ge2\sqrt{ab}\)(bđt cosi) => \(\frac{1}{a+b}\le\frac{1}{2\sqrt{ab}}=\frac{1}{2\sqrt{3}}=\frac{\sqrt{3}}{6}\)

=> \(3+\frac{1}{a+b}\le3+\frac{\sqrt{3}}{6}=\frac{18+\sqrt{3}}{6}\) => \(M\le\frac{18+\sqrt{3}}{6}\)

Dấu "=" xảy ra<=> \(\hept{\begin{cases}a=b\\ab=3\end{cases}}\) <=> \(a=b=\sqrt{3}\)

Vậy MaxM = \(\frac{18+\sqrt{3}}{2}\) <=> \(a=b=\sqrt{3}\)

\(M=\frac{a^2b+b^2a+1}{a+b}=ab+\frac{1}{a+b}\le ab+\frac{1}{2\sqrt{ab}}=3+\frac{1}{\sqrt{3}}\)(BĐT tương đương)

dấu "=" xảy ra khi và chỉ khi \(a=b=1\)

\(< =>MIN:M=3+\frac{1}{\sqrt{3}}\)

Ta có: 

\(A=\left(1+tan^2x\right)cos^2x-\left(1+cot^2x\right)\left(cos^2x-1\right)\)

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

\(=1-1=0\)

\(B=tan72^o-cot18^o+sin^230^o+sin^260^o\)

\(=tan72^o-tan72^o+sin^230^o+cos^230^o\)

\(=1\)