\(C=sin^4a\left(3-2sin^2a\right)+cos^4a\left(3-2cos^2a\right)\)

\(=sin^4a\left(1+2cos^2a\right)+cos^4a\left(1+2sin^2a\right)\)

\(=sin^4a+cos^4a+2sin^2a.cos^2a\left(sin^2a+cos^2a\right)\)

\(=sin^4a+cos^4a+2sin^2a.cos^2a=\left(sin^2a+cos^2a\right)^2=1\)

\(B=\left(3sina+4cosa\right)^2+\left(4sina-3cosa\right)^2\)

\(=9sin^2a+24sina.cosa+16cos^2a+16sin^2a-24sina.cosa+9cos^2a\)

\(=33sin^2a+33cos^2a=33\)

\(\sqrt{n^2+n^2\left(n+1\right)^2+\left(n+1\right)^2}\)

\(=\sqrt{n^2+\left(n^2+n\right)^2+\left(n^2+2n+1\right)}\)

\(=\sqrt{2\left(n^2+n\right)+\left(n^2+n\right)^2+1}=\sqrt{\left(n^2+n+1\right)^2}\)

\(=\left|n^2+n+1\right|=n^2+n+1\) vì \(n^2+n+1=\left(n+\frac{1}{4}\right)^2+\frac{3}{4}>0\)

Do đó nếu \(\sqrt{n^2+n^2\left(n+1\right)^2+\left(n+1\right)^2}\) là số nguyên nếu n là số nguyên

Ta có:\(\frac{1}{xy}+\frac{1}{yz}+\frac{1}{zx}=1\Rightarrow x+y+z=xyz\)

Dễ có một vài phép biến đổi cơ bản và bất đẳng thức AM - GM:\(\frac{x}{\sqrt{yz\left(1+x^2\right)}}=\frac{x}{\sqrt{yz+x^2yz}}=\frac{x}{\sqrt{yz+x\left(x+y+z\right)}}=\frac{x}{\sqrt{\left(x+z\right)\left(x+y\right)}}\)

\(=\sqrt{\frac{x}{x+z}\cdot\frac{x}{x+y}}\le\frac{\frac{x}{x+z}+\frac{x}{x+y}}{2}\)

Khi đó:\(LHS\le\frac{1}{2}\left(\frac{x}{x+y}+\frac{y}{x+y}+\frac{x}{x+z}+\frac{z}{x+z}+\frac{y}{z+y}+\frac{z}{z+y}\right)=\frac{3}{2}\)

Đẳng thức xảy ra tại \(x=y=z=\sqrt{3}\)

- Biến đổi \(2+\sqrt{3}=\frac{4+2\sqrt{3}}{2}=\frac{\left(\sqrt{3}+1\right)^2}{2}\)

- Tương tự \(2-\sqrt{3}=\frac{\left(\sqrt{3}-1\right)^2}{2}\)

Vậy A \(=\frac{\left(\sqrt{3}+1\right)^2}{2\sqrt{2}+\sqrt{6}+\sqrt{2}}+\frac{\left(\sqrt{3}-1\right)^2}{2\sqrt{2}-\sqrt{6}+\sqrt{2}}\)

\(=\frac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{6}}=\sqrt{2}\)

Với a,b > = 0 và a + b = a²b²

Ta có:

\(VT=\sqrt{a+b+4\sqrt{a+b+2ab+1}}=\sqrt{a^2b^2+4\sqrt{a^2b^2+2ab+1}}\)

\(=\sqrt{a^2b^2+4\sqrt{\left(ab+1\right)^2}}=\sqrt{a^2b^2+4\left(ab+1\right)}\)

\(=\sqrt{a^2b^2+4ab+4}=\sqrt{\left(ab+2\right)^2}=ab+2=VP\)

=> đpcm