Question

1. Tìm giá trị nhỏ nhất của các đa thức :

a, x² - 2x + 5.

b, Q= 2x² - 6x.

c, M= x² + y² - x + 6y+10.

Answer 1

\(a.A=x^2-2x+5=\left(x-1\right)^2+4\) ≥ 4

⇒ \(A_{Min}=4\) ⇔ \(x=1\)

\(b.Q=2x^2-6x=2\left(x^2-2.\dfrac{3}{2}+\dfrac{9}{4}\right)-\dfrac{9}{2}=2\left(x-\dfrac{3}{2}\right)^2-\dfrac{9}{2}\) ≥ \(-\dfrac{9}{2}\)

⇒ \(Q_{Min}=-\dfrac{9}{2}\) ⇔ \(x=\dfrac{3}{2}\)

\(c.M=x^2+y^2-x+6y+10=x^2-2.\dfrac{1}{2}x+\dfrac{1}{4}+y^2+6y+9+1-\dfrac{1}{4}=\left(x-\dfrac{1}{2}\right)^2+\left(y+3\right)^2+\dfrac{3}{4}\) ≥ \(\dfrac{3}{4}\)

⇒ \(M_{Min}=\dfrac{3}{4}\) ⇔ \(x=\dfrac{1}{2};y=-3\)