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$L=\sqrt{x^2+2y^2-6x+4y+11}+\sqrt{x^2+3y^2+2x+6y+4} \\\Leftrightarrow L=\sqrt{(x-3)^2+2(y+1)^2}+\sqrt{(x+1)^2+3(y+1)^2} \\ \Rightarrow L \ge \sqrt{(x-3)^2}+\sqrt{(x+1)^2} \\ \Leftrightarrow L \ge |3-x|+|x+1| \\ \Rightarrow L \ge 4 \ \ \ \ \ \ \ \text{Dấu đẳng thức $\Leftrightarrow \begin{cases}x=2\\y=-1 \end{cases}$}$
 
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