Question

[Hoàng Như Yến]

M = 2x²+3y²+x-y-3

giúp mình nha!!!!!!!!!!

Answer 1

[Nguyễn Minh Đăng]

Tìm GTNN??

Ta có:

\(M=2x^2+3y^2+x-y-3\)

\(M=2\left(x^2+\frac{1}{2}x+\frac{1}{16}\right)+3\left(y^2-\frac{1}{3}y+\frac{1}{36}\right)-\frac{77}{24}\)

\(M=2\left(x+\frac{1}{4}\right)^2+3\left(y-\frac{1}{6}\right)^2-\frac{77}{24}\ge-\frac{77}{24}\left(\forall x,y\right)\)

Dấu "=" xảy ra khi: \(\hept{\begin{cases}2\left(x+\frac{1}{4}\right)^2=0\\3\left(y-\frac{1}{6}\right)^2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{4}\\y=\frac{1}{6}\end{cases}}\)

Vậy \(Min_M=-\frac{77}{24}\Leftrightarrow\hept{\begin{cases}x=-\frac{1}{4}\\y=\frac{1}{6}\end{cases}}\)

Answer 2

[l҉o҉n҉g҉ d҉z҉]

Tìm Min ?

M = 2x² + 3y² + x - y - 3

= 2( x² + 1/2x + 1/16 ) + 3( y² - 1/3y + 1/36 ) - 77/24

= 2( x + 1/4 )² + 3( y - 1/6 )² - 77/24

\(\hept{\begin{cases}2\left(x+\frac{1}{4}\right)^2\ge0\forall x\\3\left(y-\frac{1}{6}\right)^2\ge0\forall y\end{cases}}\Rightarrow2\left(x+\frac{1}{4}\right)^2+3\left(y-\frac{1}{6}\right)^2-\frac{77}{24}\ge-\frac{77}{24}\forall x,y\)

Dấu "=" xảy ra <=> x = -1/4 ; y = 1/6

=> MinM = -77/24 <=> x = -1/4 ; y = 1/6