\(Q=-2\left(x-\dfrac{3}{2}\right)^2+\dfrac{25}{2}\le\dfrac{25}{2}\)

\(Q_{max}=\dfrac{25}{2}\) khi \(x=\dfrac{3}{2}\)

\(A=\dfrac{9\left(x^2+2\right)-9x^2+6x-1}{x^2+2}=9-\dfrac{\left(3x-1\right)^2}{x^2+2}\le9\)

\(A_{max}=9\) khi \(x=\dfrac{1}{3}\)

\(A=\dfrac{12x+34}{2\left(x^2+2\right)}=\dfrac{-\left(x^2+2\right)+x^2+12x+36}{2\left(x^2+2\right)}=-\dfrac{1}{2}+\dfrac{\left(x+6\right)^2}{2\left(x^2+2\right)}\le-\dfrac{1}{2}\)

\(A_{min}=-\dfrac{1}{2}\) khi \(x=-6\)

A= (x²-2xy +y²)+(2x-2y)+1+(y²-8y+16)

A= (x-y)² +2(x-y) +1 +(y-4)²

A= (x-y+1)² +(y-4)²

Vì (x-y+1)² +(y-4)² >= 0 với mọi x,y

Dấu = xảy ra <=> x-y+1=0 và y-4=0

                   <=> x=3 và y=4

A = /2*-5-3/1+/2*-5

a) x² - 2x + 5 = (x - 1)² + 4 >= 4

Min là 4 khi x = 1

A=[(X²-2XY+Y²)+2(X-Y)+1]+(Y²-8Y+16)=(X-Y+1)²+(Y-4)²>=0

=>Amin=0 khi y=4;x=3