Question

Tìm GTNN của BT:

\(2x^2+4y^2+4xy+2x+4y+9\)

Answer 1

=x² +4y²+4xy+1+2x+4y+x²+9

=(x+2y)²+2(x+2y)+1+x²+9

=(x+2y+1)²+x²+9

có (x+2y+1)²≥0 với mọi x,y

x²≥0 với mọi x

⇒(x+2y+1)²+x² ≥0với mọi x,y

⇒(x+2y+1)²+x²+9≥9với mọi x,y

⇒

Answer 2

ta có :

A = 2x²+4y²+4xy+2x+4y+9 = 2x²+2x+4y²+4y+4xy+9

= 2x(x+1)+4y(y+1)+4xy+9

= 2x(x+1)+4y(y+x+1)+9

= (x+1)(2x+4y²)+9

=> A lớn hơn hoặc bằng 9

=> min A là 9

Answer 3

\(2x^2+4y^2+4xy+2x+4y+9\\ =x^2+x^2+4y^2+4xy+2x+4y+1+8\\ =\left(x^2+4xy+4y^2\right)+\left(2x+4y\right)+1+x^2+8\\ =\left(x+2y\right)^2+2\left(x+2y\right)+1+x^2+8\\ =\left[\left(x+2y\right)^2+2\left(x+2y\right)+1\right]+x^2+8\\ =\left(x+2y+1\right)^2+x^2+8\\ Do\text{ }x^2\ge0\forall x\\ \left(x+2y+1\right)^2\ge0\forall x;y\\ \Rightarrow\left(x+2y+1\right)^2+x^2\ge0\forall x;y\\ \Rightarrow\left(x+2y+1\right)^2+x^2+8\ge8\forall x;y\\ Dấu\text{ }"="\text{ }xảy\text{ }ra\text{ }khi:\left\{{}\begin{matrix}x^2=0\\\left(x+2y+1\right)^2=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x+2y+1=0\end{matrix}\right.\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\2y+0+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\2y=-1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right.\\ Vậy\text{ }GTNN\text{ }của\text{ }biểu\text{ }thức\text{ }là\text{ }8\text{ }khi\text{ }\left\{{}\begin{matrix}x=0\\y=-\dfrac{1}{2}\end{matrix}\right.\)