Ta có : \(\left(2^{10}+1\right)^{2010}=\left(1024+1\right)^{2010}=1025^{2010}=\left(25.41\right)^{2010}=25^{2010}.41^{2010}\)

Thấy thấy \(25^{2010}⋮25^{2010}\Rightarrow25^{2010}.41^{2010}⋮25^{2010}\)hay \(\left(2^{10}+1\right)^{2010}⋮25^{2010}\)

Theo BĐT \(\frac{1}{a}+\frac{1}{b}\ge\frac{4}{a+b}\) ta có:

\(\frac{ab}{c+1}=\frac{ab}{a+c+b+c}\le\frac{1}{4}\left(\frac{ab}{a+c}+\frac{ab}{b+c}\right)\)

Tương tự ta cũng có các BĐT sau:

\(\frac{bc}{a+1}\le\frac{1}{4}\left(\frac{bc}{a+b}+\frac{bc}{a+c}\right);\frac{ca}{b+1}\le\frac{1}{4}\left(\frac{ac}{a+b}+\frac{ac}{b+c}\right)\)

Cộng theo vế các BĐT cùng dấu có:

\(Q\le\frac{1}{4}\left(\frac{c\left(a+b\right)}{a+b}+\frac{a\left(b+c\right)}{b+c}+\frac{b\left(c+a\right)}{c+a}\right)\)

\(=\frac{1}{4}\left(a+b+c\right)=\frac{1}{4}\left(a+b+c=1\right)\)

Khi a=b=c=1/3

B=\(\sqrt{17}+\sqrt{5}+1\)>\(\sqrt{16}+\sqrt{4}+1\)=4+2+1=7=\(\sqrt{49}\)>\(\sqrt{45}\)

Vậy B>C

a)1-1+1=1

a.1

b.1,140119483

c.0,353338015

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

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

\(=3-\sqrt{3}\)

=\(\sqrt{3}-1+2-\) \(\sqrt{3}=1\)

b.=\(\frac{2+\sqrt{3}-2+\sqrt{3}}{2^2-3}=2\sqrt{3}\)