\(\left(\frac{2}{5}\sqrt{16}+2\sqrt{\frac{16}{25}}\right):2\sqrt{\frac{1}{16}}=\left(\frac{2}{5}.\sqrt{4^2}+2\sqrt{\frac{4^2}{5^2}}\right):\frac{2}{\sqrt{4^2}}\)

\(=\left(\frac{2}{5}.4+2.\frac{4}{5}\right).2=\left(\frac{8}{5}+\frac{8}{5}\right).2=\frac{32}{5}\)

\(\left(\frac{2}{5}.\sqrt{16}+2\sqrt{\frac{16}{25}}\right):2\sqrt{\frac{1}{16}}\)

\(=\left(\frac{2}{5}.4+2.\frac{4}{5}\right):2.\frac{1}{4}\)

\(=\left(\frac{8}{5}+\frac{8}{5}\right):\frac{1}{2}\)

\(=\frac{16}{5}:\frac{1}{2}\)

\(=\frac{32}{5}\)

 ^...^ ^_^

\(\left(\frac{2}{5}\sqrt{16}+2\sqrt{\frac{16}{25}}\right):2\sqrt{\frac{1}{16}}\)

\(=\left(\frac{2}{5}.4+2.\frac{4}{5}\right):2.\frac{1}{4}\)

\(=\left(\frac{8}{5}+\frac{8}{5}\right):\frac{1}{2}\)

\(=\frac{16}{5}:\frac{1}{2}\)

\(=\frac{16}{5}.2\)

\(=\frac{32}{5}\)

a) = \(\frac{7}{2}\)

b) = \(\frac{643}{64}\)

c) = 0

Cách khác: Sử dụng khai triển Taylor-Maclaurin

\(\cos x=1-\dfrac{x^2}{2!}+o\left(x^2\right)\)

\(\cos2x=1-\dfrac{4x^2}{2!}+o\left(x^2\right)\)

\(\cos4x=1-\dfrac{16x^2}{2!}+o\left(x^2\right)\)

\(\Rightarrow1-\cos x\cos2x\cos4x=1-\left(1-\dfrac{x^2}{2}+o\left(x^2\right)\right)\left(1-2x^2+o\left(x^2\right)\right)\left(1-8x^2+o\left(x^2\right)\right)\)

\(=1-\left(1-\dfrac{5x^2}{2}+o\left(x^2\right)\right)\left(1-8x^2+o\left(x^2\right)\right)=1-1+8x^2+\dfrac{5x^2}{2}+o\left(x^2\right)=\dfrac{21x^2}{2}+o\left(x^2\right)\)

\(\Rightarrow\lim\limits_{x\rightarrow0}\dfrac{1-\cos x\cos2x\cos4x}{1-\cos2x}=\lim\limits_{x\rightarrow0}\dfrac{\dfrac{21}{2}x^2+o\left(x^2\right)}{2x^2+o\left(x^2\right)}=\dfrac{21}{4}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{1-cosx+\left(1-cos2x\right)cosx+\left(1-cos4x\right)cosx.cos2x}{1-cos2x}\)

\(=\lim\limits_{x\rightarrow0}\dfrac{1-cosx}{1-cos2x}+\lim\limits_{x\rightarrow0}\dfrac{\left(1-cos2x\right)}{1-cos2x}.cosx+\lim\limits_{x\rightarrow0}\dfrac{\left(1-cos4x\right)}{1-cos2x}.cosx.cos2x\)

\(=\lim\limits_{x\rightarrow0}\dfrac{1-cosx}{2\left(1-cosx\right)\left(1+cosx\right)}+\lim\limits_{x\rightarrow0}\left(cosx\right)+\lim\limits_{x\rightarrow0}\dfrac{2\left(1-cos2x\right)\left(1+cos2x\right)}{1-cos2x}.cosx.cos2x\)

\(=\lim\limits_{x\rightarrow0}\left(\dfrac{1}{2\left(1+cosx\right)}\right)+\lim\limits_{x\rightarrow0}\left(cosx\right)+\lim\limits_{x\rightarrow0}2cosx.cos2x.\left(1+cos2x\right)\)

\(=\dfrac{1}{4}+1+4=\dfrac{21}{4}\)

\(1-cosx.cos2x.cos3x...cosnx\)

\(=1-cosx+cosx\left(1-cos2x\right)+cosx.cos2x\left(1-cos3x\right)+...+cosx.cos2x...cos\left(n-1\right)x\left(1-cosnx\right)\)

Hay nói chung là:

\(1-a.b.c.d...m.n\)

\(=1-a+a\left(1-b\right)+ab\left(1-c\right)+abc\left(1-d\right)+...+ab...m\left(1-n\right)\)

Học dỏi nha :)) 
~ Good luck ~

\(\sqrt{\frac{289+4\sqrt{72}}{16}}+\sqrt{\frac{129}{16}+\sqrt{2}}\)

\(=\sqrt{\frac{288+2\times12\sqrt{2}+1}{4^2}}+\sqrt{\frac{128+2\sqrt{12}+1}{4^2}}\)

\(=\sqrt{\frac{\left(\sqrt{288}+1\right)^2}{4^2}}+\sqrt{\frac{\left(\sqrt{128}+1\right)^2}{4^2}}\)

\(=\frac{\sqrt{288}+1}{4}+\frac{\sqrt{128}+1}{4}\)

\(=\frac{12\sqrt{2}+8\sqrt{2}+2}{4}\)

\(=\frac{1+10\sqrt{2}}{2}\)

\(\sqrt{16-6\sqrt{7}}=\sqrt{9-2.3\sqrt{7}+7}=\sqrt{\left(3-\sqrt{7}\right)^2}=3-\sqrt{7};\sqrt{10-2\sqrt{21}}=\sqrt{3-2\sqrt{3}\sqrt{7}+7}=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\Rightarrow\sqrt{16-6\sqrt{7}}+\sqrt{10-2\sqrt{21}}=3-\sqrt{3}\)