Question

[Anh Kiên lớp 7 Lê]

tim GTNNcua da thuc:

H=x^2+y^2-xy-x+y+1

Answer 1

[Hiền Thương]

H=x²+y²-xy-x+y+1

=>4H=4x²+4y²-4xy -4x +4y +4

4H= (4x²+y²+1-4xy-4x+2y)+3y²+2y +3

4H=(2x-y-1)²+3(y²+\(\dfrac{2}{3}y\) +\(\dfrac{1}{9}\))+ \(\dfrac{8}{3}\)

4H = (2x-y-1)²+3(y+\(\dfrac{1}{3}\) )² \(+\dfrac{8}{3}\) \(\ge\)\(\dfrac{8}{3}\)(vì \(\left(2x-y-1\right)^2\ge0\forall x,y;3\left(y+\dfrac{1}{3}\right)^2\ge0\forall y\)

=> H \(\ge\dfrac{2}{3}\) 

Dấu ''=' xảy ra khi \(\left\{{}\begin{matrix}y=-\dfrac{1}{3}\\x=\dfrac{1}{3}\end{matrix}\right.\)

Vậy Hmin= 2/3khi x=1/3;y=-1/3

Answer 2

[Yen Nhi]

\(H=x^2+y^2-xy-x+y+1\)

\(\Rightarrow4H=4x^2+4y^2-4xy-4x+4y+4\)

\(\Leftrightarrow4A=4x^2+y^2+1-4xy+2y-4x+3y^2+2y+3\)

\(\Rightarrow3.4A=3\left(2x-y-1\right)^2+9y^2+6y+9\)

\(\Leftrightarrow12A=3\left(2x-y-1\right)^2+\left(9y^2+6y+1\right)+8\)

\(\Leftrightarrow12A=3\left(2x-y-1\right)^2+\left(3y+1\right)^2+8\)

Mà \(3\left(2x-y-1\right)^2+\left(3y+1\right)^2+8\ge8\forall x,y\)

\(\Rightarrow12A\ge8\)

\(\Rightarrow A\ge\dfrac{8}{12}=\dfrac{2}{3}\)

Dấu "=" xảy ra khi \(\left\{{}\begin{matrix}2x-y-1=0\\3y+1=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=\dfrac{1}{3}\\y=-\dfrac{1}{3}\end{matrix}\right.\)