Question

[Nguyễn Minh Châu]

tìm GTNN của các biểu thức sau

c) C= 2x^2+y^2+2xy-8x+2019

d) D=(x-1)(x+2)(x+3)(x+6)

Answer 1

[@Nk>↑@]

c)\(C=2x^2+y^2+2xy-8x+2019\)

\(=\left(x^2+2xy+y^2\right)+\left(x^2-8x+16\right)+2003\)

\(=\left(x+y\right)^2+\left(x-4\right)^2+2003\)

Vì \(\left(x+y\right)^2\ge0,\forall x,y\)

\(\left(x-4\right)^2\ge0,\forall x\)

Nên \(\left(x+y\right)^2+\left(x-4\right)^2\ge0\)

\(\Rightarrow\left(x+y\right)^2+\left(x-4\right)^2+2003\ge2003\)

\(\Rightarrow min_C=2003\) khi \(\left\{{}\begin{matrix}x-4=0\\x+y=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=4\\y=-4\end{matrix}\right.\)

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

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

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

Đặt \(t=x^2+5x\) ta được:

\(D=\left(t-6\right)\left(t+6\right)=t^2-36\)

Vì \(t^2\ge0,\forall t\)

\(\Rightarrow t^2-36\ge-36\)

\(\Rightarrow min_D=-36\) khi \(t=0\)

\(\Leftrightarrow x^2-5x=0\)

\(\Leftrightarrow x\left(x-5\right)=0\)

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

Vậy Min D = -36 khi \(x\in\left\{0;5\right\}\)