Question

Cho A= x⁴ + x² - 6x + 9

tìm MinA

Answer 1

A = (x^4-2x^2+1)+(3x^2-6x+3)+5

   = (x^2-1)^2+3.(x-1)^2+5 >= 5 

Dấu "=" xảy ra <=>  x^2-1=0 và x-1=0 <=> x=1

Vậy Min A = 5 <=> x=1

k mk nha

Answer 2

A=\(x^4+x^2-6x+9\)

\(=\left(x^4-2x^2+1\right)\left(3x^2-6x+3\right)+5\)

\(=\left[\left(x^2\right)^2-2x^2.1+1^2\right]+3.\left(x^2-2x+1\right)+5\)

\(=\left(x^2-1\right)^2+3.\left(x-1\right)^2+5\ge5\)

Min A=5 khi \(\hept{\begin{cases}x^2-1=0\\x-1=0\end{cases}}\)=> x = 1