cho x^2 + y^2 =1. CMR -5<3x+4y<5

Đáp án:

 Áp dụng bđt Bunhia ta có:

$\begin{array}{l} {\left( {3x + 4y} \right)^2} \le \left( {{3^2} + {4^2}} \right)\left( {{x^2} + {y^2}} \right)\\  \Rightarrow {\left( {3x + 4y} \right)^2} \le 25.1 = 25\left( {do:{x^2} + {y^2} = 1} \right)\\  \Rightarrow  - 5 \le 3x + 4y \le 5\\ Dấu = xảy\,ra \Leftrightarrow \left\{ \begin{array}{l} \frac{x}{3} = \frac{y}{4}\\ {x^2} + {y^2} = 1 \end{array} \right. \Rightarrow \left\{ \begin{array}{l} x = \frac{3}{5}\\ y = \frac{4}{5} \end{array} \right. \end{array}$