Question

[vương đức]

Tìm GTNN của biểu thức :

A= x²-3x+1 B= x²+2y²-2xy+2x-10y+17

C= \(\dfrac{-3}{x^2-x+2}\) D= \(\dfrac{2x^2-16x+41}{x^2-8x+22}\) E= \(\dfrac{4x^2-6x+3}{\left(2x-1\right)^2}\)

Answer 1

[Nguyễn Lê Phước Thịnh]

a: \(A=x^2-3x+\dfrac{9}{4}-\dfrac{5}{4}=\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}>=-\dfrac{5}{4}\)

Dấu '=' xảy ra khi x=3/2

c: \(x^2-x+2=\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}>=\dfrac{7}{4}\)

=>\(\dfrac{3}{\left(x-\dfrac{1}{2}\right)^2+\dfrac{7}{4}}< =3:\dfrac{7}{4}=\dfrac{12}{7}\)

=>C>=-12/7

Dấu '=' xảy ra khi x=1/2