Question

Tìm Giá trị nhỏ nhất 
a) A= x2 - 2x + y2 -4y +7 
b) B= (x-1) (x+2) (x+3) ( x+6)

Answer 1

a: \(A=x^2-2x+1+y^2-4y+4+2=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\forall x,y\)

Dấu '=' xảy ra khi x=1 và y=2

b: \(B=\left(x^2+5x-6\right)\left(x^2+5x+6\right)\)

\(=\left(x^2+5x\right)^2-36\ge-36\forall x\)

Dấu '=' xảy ra khi x=0 hoặc x=-5

Answer 2

a) \(A=x^2-2x+y^2-4y+7\)

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

\(=\left(x-1\right)^2+\left(y-2\right)^2+2\ge2\)

\(A_{min}=2\Leftrightarrow x=1;y=2\).

b) \(B=\left(x-1\right)\left(x+2\right)\left(x+3\right)\left(x+6\right)\)

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

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

-Đặt \(t=x^2+5x-6\)

\(B=t\left(t+12\right)=t^2+12t=t^2+12t+36-36=\left(t+6\right)^2-36\ge-36\)

\(B_{min}=-36\Leftrightarrow t=-6\Leftrightarrow x^2+5x-6=-6\Leftrightarrow x^2+5x=0\Leftrightarrow x\left(x+5\right)=0\Leftrightarrow x=0hayx=-5\)