[Toán 11]Chứng minh

N

nerversaynever

[TEX]\begin{array}{l}f\left( x \right) = a{x^2} + bx - \left( {\frac{a}{2} + \frac{{2b}}{3}} \right) = 0\\f\left( 0 \right)f\left( {\frac{3}{4}} \right) = - \left( {\frac{a}{2} + \frac{{2b}}{3}} \right)\left( {\frac{a}{{16}} + \frac{b}{{12}}} \right) = - \frac{1}{8}{\left( {\frac{a}{2} + \frac{{2b}}{3}} \right)^2} \le 0 \to dpcm\end{array}[/TEX]
 
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