Question

tìm min:

G=x²+3y²+2xy-6y+3

các nạn giúp mình với!!!!!!!!!!!!!!!!!!!!

Answer 1

Ta có:

\(G=x^2+3y^2+2xy-6y+3\)

\(G=\left(x^2+2xy+y^2\right)+\left(2y^2-6y+\frac{18}{4}\right)-\frac{3}{2}\)

\(G=\left(x+y\right)^2+2\left(y-\frac{3}{2}\right)^2-\frac{3}{2}\ge-\frac{3}{2}\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+y\right)^2=0\\2\left(y-\frac{3}{2}\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{3}{2}\end{cases}}\)

Vậy Min(G) = -3/2 khi \(\hept{\begin{cases}x=-\frac{3}{2}\\y=\frac{3}{2}\end{cases}}\)

Answer 2

G = x² + 3xy² + 2xy - 6y + 3

<=> G = ( x² + 2xy + y² ) + ( y² - 6y + 9 ) - 6

<=> G = ( x + y )² + ( y - 3 )² - 6

Vì ( x + y )²\(\ge\)0 ; ( y - 3 )²\(\ge\)0\(\forall\)x ; y

=> G = ( x + y )² + ( y - 3 )² - 6\(\ge\)- 6

Dấu "=" xảy ra <=>\(\orbr{\begin{cases}\left(x+y\right)^2=0\\\left(y-3\right)^2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-y\\y=3\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=-3\\y=3\end{cases}}\)

Vậy minG = - 6 <=> x = - 3 ; y = 3

Answer 3

Xin lỗi tớ nhầm ) Đăng làm đúng nhé