[Toán 8] biểu thức đại số

V

vipboycodon

A = $x^2+3x-10$
= $x^2+2x.\dfrac{3}{2}+\dfrac{9}{4}-\dfrac{49}{4}$
= $(x+\dfrac{3}{2})^2-\dfrac{49}{4} \ge -\dfrac{49}{4}$
Vậy Min A = $\dfrac{-49}{4}$ khi $x = -\dfrac{3}{2}$
 
V

vipboycodon

B = $3x^2-2x-5$
= $3(x^2-\dfrac{2x}{3}-\dfrac{5}{3})$
= $3(x^2-2x.\dfrac{1}{3}+\dfrac{1}{9}-\dfrac{16}{9})$
= $3(x-\dfrac{1}{3})^2-\dfrac{16}{3} \ge \dfrac{-16}{3}$
Vậy Min B = $\dfrac{-16}{3}$ khi $x = \dfrac{1}{3}$
 
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