Question

Giải bài 9 hộ mình

Answer 1

a/ \(x^2+2x+3=\left(x^2+2x+1\right)+2=\left(x+1\right)^2+2\ge2>0\) với mọi số thực x

b/ \(A=\left(x-1\right)\left(x-3\right)+11=x^2-4x+14=\left(x^2-4x+4\right)+10=\left(x-2\right)^2+10\ge10\)

Suy ra Min A = 10 <=> x = 2

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

Đặt \(t=x^2+3x\)  thì \(B=t^2-1\ge-1\)

Do đó Min B = -1 <=> t = 0 <=> \(\left[\begin{array}{nghiempt}x=0\\x=-3\end{array}\right.\)

c/\(C=5-4x^2+4x=-\left(4x^2-4x+1\right)+6=-\left(2x-1\right)^2+6\le6\)

Suy ra Max C = 6 <=> x = 1/2

\(D=-x^2-4x-y^2+2y=-\left(x^2+4x+4\right)-\left(y^2-2y+1\right)+5\)

\(=-\left(x+2\right)^2-\left(y-1\right)^2+5\le5\)

Suy ra Max D = 5 <=> (x;y) = (-2;1)