\(A=x^2+3xy+6x+5y^2+7y-2\)
\(=\left[x^2+2x\left(3+\dfrac{3}{2}y\right)+\left(3+\dfrac{3}{2}y\right)^2\right]+5y^2+7y-2-\left(3+\dfrac{3}{2}y\right)^2\)\(=\left(x+3+\dfrac{3}{2}y\right)^2+5y^2+7y-2-9-9y-\dfrac{9}{4}y^2\)\(=\left(x+3+\dfrac{3}{2}y\right)^2+\dfrac{11}{4}y^2-2y-11\)
\(=\left(x+3+\dfrac{3}{2}\right)^2+\dfrac{11}{4}\left(y^2-\dfrac{8}{11}y+\dfrac{16}{121}\right)-\dfrac{125}{11}\)\(=\left(x+3+\dfrac{3}{2}y\right)^2+\dfrac{11}{4}\left(x-\dfrac{4}{11}\right)^2-\dfrac{125}{11}\ge\dfrac{-125}{11}\)Vậy \(Min_A=\dfrac{-125}{11}\) khi \(\left[{}\begin{matrix}x+3+\dfrac{3}{2}y=0\\x-\dfrac{4}{11}=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-\dfrac{74}{33}\\x=\dfrac{4}{11}\end{matrix}\right.\)
Biết số nhọ nhưng vẫn làm tiếp:)
\(2,x^4+3x^2+2x+2=\left(x^4+2x^2+1\right)+\left(x^2+2x+1\right)=\left(x^2+1\right)^2+\left(x+1\right)^2>0\left(đpcm\right)\)
\(b,x^2+y^2+z^2+xy+yz+zx\ge0\)
\(\Leftrightarrow2\left(x^2+y^2+z^2+xy+yz+zx\right)\ge0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+2xz+z^2\right)+\left(y^2+2yz+z^2\right)\ge0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+z\right)^2+\left(y+z\right)^2\ge0\)
Đúng với mọi x , y ,z
c,\(x^2+y^2+xy+x+y+1\ge0\)
\(\Leftrightarrow2\left(x^2+y^2+xy+y+x+1\right)\ge0\)
\(\Leftrightarrow\left(x^2+2xy+y^2\right)+\left(x^2+2x+1\right)+\left(y^2+2y+1\right)\ge0\)
\(\Leftrightarrow\left(x+y\right)^2+\left(x+1\right)^2+\left(y+1\right)^2\ge0\)
Đúng với mọi x , y
Nhác làm quá!!! Gõ được mấy dòng r mà nó tự nhiên quay trở về hết :( Tức lắm luôn^^!....Số nhọ!:(