toán 8: tìm min , max

V

vipboycodon

$A = \dfrac{3-4x}{x^2+1}$
= $\dfrac{x^2-4x+4-x^2-1}{x^2+1}$
= $\dfrac{(x-2)^2-(x^2+1)}{x^2+1}$
= $\dfrac{(x-2)^2}{x^2+1}-1 \ge -1$
Min A = -1 khi $x = 2$
 
V

vipboycodon

$A = \dfrac{3-4x}{x^2+1}$
= $\dfrac{4x^2+4-4x^2-4x-1}{x^2+1}$
= $\dfrac{4(x^2+1)-(4x^2+4x+1)}{x^2+1}$
= $4-\dfrac{(2x+1)^2}{x^2+1} \le 4$
Max A = 4 khi $x = \dfrac{-1}{2}$
 
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