\(N=2x^2+y^2+2xy-4x-2y+3\)
\(N=\left(x^2+2xy+y^2\right)+x^2-4x-2y+3\)
\(N=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]+\left(x^2-2x+1\right)+1\)
\(N=\left(x+y-1\right)^2+\left(x-1\right)^2+1\)
Mà \(\left(x+y-1\right)\ge0\forall x;y\)
\(\left(x-1\right)^2\ge0\forall x\)
\(\Rightarrow N\ge1\)
Dấu "=" xảy ra khi :
\(\hept{\begin{cases}x+y-1=0\\x-1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}y=0\\x=1\end{cases}}\)
Vậy \(N_{Min}=1\Leftrightarrow\left(x;y\right)=\left(1;0\right)\)
\(N=2x^2+y^2+2xy-4x-2y\)\(+3\)
\(=\left(x^2+2xy+y^2\right)+x^2-2\left(2x+y\right)+3\)
\(=\left[\left(x+y\right)^2-2\left(2x+y\right)+1\right]+2+x^2\)
\(=\left(x+y+1\right)^2+x^2+2\)
\(Do\)\(\left(x+y+1\right)^2\)\(\ge\)\(0\)\(\forall\)\(x\)\(;\)\(y\)
\(x^2\)\(\ge\)\(0\)\(\forall\)\(x\)
=.>\(\left(x+y+1\right)^2+x^2+2\)\(\ge\)\(2\)\(\forall\)\(x\)\(;\)\(y\)
=>\(N\)\(\ge\)\(2\)\(\forall\)\(x\)\(;\)\(y\)
Dấu = xảy ra khi:
\(\hept{\begin{cases}\left(x+y+1\right)^2=0\\x^2=0\end{cases}}\)
=>\(\hept{\begin{cases}x+y+1=0\\x=0\end{cases}}\)
=>\(\hept{\begin{cases}x+y=-1\\x=0\end{cases}}\)
=>\(\hept{\begin{cases}y=-1\\x=0\end{cases}}\)
Vậy \(N_{min}\)\(=\)\(2\)khi \(y=-1\)\(;\)\(x=0\)
Chúc pạn họk tốt~~~!!! :3
@jiyoonmin : bn lm như cak mk á
@jiyoonmin : sai ngay từ dòng thứ 3 kìa cưng :)
\(\left[\left(A+B\right)-C\right]^2=\left(A+B\right)^2-2\left(A+B\right).C+C^2\)
chứ éo phải là \(=\left(A+B\right)^2-2\left(2A+B\right).C+C^2\)nhá :)
bn guiltykamikk đz lm đúng mak -_-
thak loz nào tk sai lắm z -_-
đm đm đm đm đm đm >-< cc chúng m ns cho bố m bk sai chỗ nào >-<
