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\(A=x^2+y^2-xy-x+y+1\)
\(12A=12x^2+12y^2-12xy-12x+12y+12\)
\(=3\left(x^2+2xy+y^2\right)+9x^2+9y^2+4-18xy-12x+12y+8\)
\(=3\left(x+y\right)^2+\left(3x-3y-2\right)^2+8\ge8\)
Dấu \(=\)khi \(\hept{\begin{cases}x+y=0\\3x-3y-2=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=\frac{1}{3}\\y=-\frac{1}{3}\end{cases}}\)
Vậy \(minA=\frac{2}{3}\).
![](https://rs.olm.vn/images/avt/0.png?1311)
\(M=2x^2+2y^2-2xy-2x+2y+2\)
\(=\left[\left(x^2-2xy+y^2\right)-\frac{4}{3}\left(x-y\right)+\frac{4}{9}\right]+\left(x^2-\frac{2}{3}x+\frac{1}{9}\right)+\left(y^2+\frac{2}{3}y+\frac{1}{9}\right)+\frac{4}{3}\)
\(=\left(x-y-\frac{2}{3}\right)^2+\left(x-\frac{1}{3}\right)^2+\left(y+\frac{1}{3}\right)^2+\frac{4}{3}\ge\frac{4}{3}\)
\(\Rightarrow M\ge\frac{2}{3}\)
\(M=x^2+y^2-xy-x+y+1\)
\(4M=4x^2+4y^2-4xy-4x+4y+1\)
\(4M=\left(4x^2-4xy+y^2\right)+3y^2-4x+4y+1\)
\(4M=\left[\left(2x-y\right)^2-2\left(2x-y\right)+1\right]+3\left(y^2+2y+1\right)-3\)
\(4M=\left(2x-y-1\right)^2+3\left(y+1\right)^2-3\)
Mà : \(\left(2x-y-1\right)^2\ge0\forall x;y\)
\(\left(y+1\right)^2\ge0\forall y\Rightarrow3\left(y+1\right)^2\ge0\forall y\)
\(\Rightarrow4M\ge-3\)
\(\Leftrightarrow M\ge-\frac{3}{4}\)
Dấu " = " xảy ra khi :
\(\hept{\begin{cases}2x-y-1=0\\y+1=0\end{cases}}\Leftrightarrow\hept{\begin{cases}2x-y=1\\y=-1\end{cases}}\Leftrightarrow\hept{\begin{cases}x=0\\y=-1\end{cases}}\)
Vậy \(M_{Min}=-\frac{3}{4}\Leftrightarrow\left(x;y\right)=\left(0;-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Đặt \(B=\frac{\left(x+y+1\right)^2}{xy+x+y}\)
Ta có bđt sau \(\left(a+b+c\right)^2\ge3\left(ab+bc+ac\right)\) tự chứng mình nha
Áp dụng \(a=x,b=y,c=1\)
Ta có : \(B=\frac{\left(x+y+1\right)^2}{xy+x+y}\ge3\)
Ta có : \(A=\frac{1}{B}+B=\frac{1}{B}+\frac{B}{9}+\frac{8B}{9}\ge\frac{2}{3}+\frac{8}{3}=\frac{10}{3}\)
Dấu " = " xảy ra khi \(x=y=1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
bài 2 nhân p vs x+y+xy rồi t định áp dụng bđt (x+y+z)(1/x+1/y+1/z)>=9 nhưng vướng