Question

[Nguyen Thi Ngoc Lan]

tìm GTNN của các BT sau

a) A=(x-1)(x+2)(x+3)(x+6)

b) B=x^2-2x+y^2+4y+8

c) C=x^2-4x+y^2-8y+6

Answer 1

[Phong Thần]

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

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

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

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

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

Vì \(\left(x^2+5x\right)^2\ge0\) với mọi x

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

\(\Rightarrow Amin=-36\Leftrightarrow x^2+5x=0\)

\(\Rightarrow x\left(x+5\right)=0\)

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=-5\end{matrix}\right.\)

b) \(B=x^2-2x+y^2+4y+8\)

\(B=\left(x^2-2x+1\right)+\left(y^2+4y+4\right)+3\)

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

Vì \(\left(x-1\right)^2\ge0\) với mọi x

\(\left(y+2\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-1\right)^2+\left(y+2\right)^2\ge0\) với mọi x,y

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

\(\Rightarrow Bmin=3\Leftrightarrow\left\{{}\begin{matrix}x=1\\y=-2\end{matrix}\right.\)

c) \(C=x^2-4x+y^2-8y+6\)

\(C=\left(x^2-4x+4\right)+\left(y^2-8y+16\right)-14\)

\(C=\left(x-2\right)^2+\left(y-4\right)^2-14\)

Vì \(\left(x-2\right)^2\ge0\) với mọi x

\(\left(y-4\right)^2\ge0\) với mọi y

\(\Rightarrow\left(x-2\right)^2+\left(y-4\right)^2-14\ge-14\) với mọi x,y

\(\Rightarrow Cmin=-14\Leftrightarrow\left\{{}\begin{matrix}x=2\\y=4\end{matrix}\right.\)