Cho a, b là hai số thực, x và y là hai số thực dương.Chứng minh rằng: \(\frac{{{{\rm{a}}^2}}}{{\rm{x}}} + \frac{{{{\rm{b}}^2}}}{{\rm{y}}} \ge \frac{{{{({\rm{a  + }}{\rm{ b)}}}^2}}}{{{\rm{x  +  y}}}}\)

Lời giải:

\(\begin{array}{l}2x + 3 = 10x - 4x - 9\\2x - 10x + 4x =  - 9 - 3\\ - 4x =  - 12\\x = 3\end{array}\)

Vậy x = 3.

Lời giải:

\(\begin{array}{l}\frac{3}{{35}} + \frac{3}{{63}} + \frac{3}{{99}} + ... + \frac{3}{{x(x + 2)}} = \frac{{24}}{{35}}\\\frac{3}{{5.7}} + \frac{3}{{7.9}} + \frac{3}{{9.11}} + ... + \frac{3}{{x(x + 2)}} = \frac{{24}}{{35}}\\\frac{3}{2}.\left( {\frac{2}{{5.7}} + \frac{2}{{7.9}} + ... + \frac{2}{{x(x + 2)}}} \right) = \frac{{24}}{{35}}\\\frac{3}{2}.\left( {\frac{1}{5} - \frac{1}{7} + \frac{1}{7} - \frac{1}{9} + ... + \frac{1}{x} - \frac{1}{{x + 2}}} \right) = \frac{{24}}{{35}}\\\frac{3}{2}.\left( {\frac{1}{5} - \frac{1}{{x + 2}}} \right) = \frac{{24}}{{35}}\\\frac{1}{5} - \frac{1}{{x + 2}} = \frac{{16}}{{35}}\\\frac{1}{{x + 2}} = \frac{1}{5} - \frac{{16}}{{35}}\\\frac{1}{{x + 2}} = \frac{{ - 9}}{{35}}\\x = \frac{{ - 53}}{9}\end{array}\)