So sánh và chứng minh các biểu thức sau

a. 

So sánh: \(\sqrt {17} + \sqrt {26} + 1\) và \(\sqrt {99}\)     

Ta có: \(\sqrt {17} > \sqrt {16} ;\sqrt {26} > \sqrt {25} => \sqrt {17} + \sqrt {26} + 1 >\sqrt {16} + \sqrt {25} + 1 = 4 + 5 + 1 = 10\)

Mà \(10 = \sqrt {100} > \sqrt {99}\)

Vậy: \(\sqrt {17} + \sqrt {26} + 1 > \sqrt {99}\) .   

b.

Chứng minh: \(\frac{1}{{\sqrt 1 }} + \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 3 }} + .... + \frac{1}{{\sqrt {99} }} + \frac{1}{{\sqrt {100} }}\, > 10\)

Ta có: \(\frac{1}{{\sqrt 1 }} > \frac{1}{{\sqrt {100} }};\frac{1}{{\sqrt 2 }} > \frac{1}{{\sqrt {100} }};\frac{1}{{\sqrt 3 }} > \frac{1}{{\sqrt {100} }};...;\frac{1}{{\sqrt {99} }} > \frac{1}{{\sqrt {100} }}\)

Suy ra:  \(\frac{1}{{\sqrt 1 }} + \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 3 }} + .... + \frac{1}{{\sqrt {100} }}\, > 100.\frac{1}{{\sqrt {100} }} = 10 \)

Vậy: \( \frac{1}{{\sqrt 1 }} + \frac{1}{{\sqrt 2 }} + \frac{1}{{\sqrt 3 }} + .... + \frac{1}{{\sqrt {100} }}\, > 10\)

c.

Ta có: \(P = \frac{1}{{1008}} + \frac{1}{{1009}} + \frac{1}{{1010}} + ... + \frac{1}{{2014}} + \frac{1}{{2015}}\)

\(= \left( {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{1006}} + \frac{1}{{1007}} + \frac{1}{{1008}} + ... + \frac{1}{{2014}} + \frac{1}{{2015}}} \right) - \left( {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{1006}} + \frac{1}{{1007}}} \right)\)

\(= \left( {1 + \frac{1}{2} + \frac{1}{3} + ... + \frac{1}{{1006}} + \frac{1}{{1007}} + \frac{1}{{1008}} + ... + \frac{1}{{2014}} + \frac{1}{{2015}}} \right) - 2\left( {\frac{1}{2} + \frac{1}{4} + \frac{1}{6} + ... + \frac{1}{{2012}} + \frac{1}{{2014}}} \right)\)

\(= 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ...... + \frac{1}{{2013}} - \frac{1}{{2014}} + \frac{1}{{2015}}= S.\)

Do đó \({\left( {S - P} \right)^{2016}}= 0\)