Question

tìm Min:
\(A=2x^2-4x+5\)

Answer 1

`A = 2(x^2 - 2x + 5/2)`

`= 2(x^2 - 2x + 1 + 3/2)`

`= 2(x-1)^2 + 3 >= 0 + 3 = 3`

Dấu bằng xảy  ra `<=> x = 1`

Answer 2

\(=2\left(x^2-2x+\dfrac{5}{2}\right)\)

\(=2\left(x^2-2x+1+\dfrac{3}{2}\right)\)

\(=2\left(x-1\right)^2+3\ge3\)

Dấu '=' xảy ra khi x=1