[Toán 8]Ôn tập học kì

M

minhduccay

Last edited by a moderator:
V

vipboycodon

1.
$\dfrac{x^2-4x+1}{x^2}$
= $1-\dfrac{4}{x}+\dfrac{1}{x^2}$
= $\dfrac{1}{x^2}-\dfrac{4}{x}+4-3$
= $(\dfrac{1}{x}-2)^2-3 \ge -3$
Vậy min = -3 khi $x = \dfrac{1}{2}$
 
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