[Toán 8] Violympic vòng 9

C

chonhoi110

Ta có: $x^2 + y^2 = 6x -5$

$\leftrightarrow x^2+y^2 - 6x +5 =0$

$\leftrightarrow (x-3)^2 + y^2 = 4$

$\rightarrow y^2 = 4 -(x-3)^2 \ge \; 0 \leftrightarrow (x-3)^2 \le \; 4$

$\leftrightarrow -2 \le \; x-3 \le \; 2 \leftrightarrow 1 \le \; x \le \; 5 $

$\leftrightarrow 6\le \; 6x \le \;30 \leftrightarrow 1 \le \; 6x-5 \le \; 25$

$\leftrightarrow 1 \le \; P \le \; 25$

Min $P=1 \leftrightarrow x=1 ; y=0$

Max $P = 25 \leftrightarrow x=5 ; y=0$
 
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