Tìm GTNN M = 5x² + y² – 4x – 2xy + 2023

2 bình luận về “Tìm GTNN M = 5x² + y² – 4x – 2xy + 2023”

1. Giải đáp:

   M = 5x^2 + y^2  -4x – 2xy + 2023

   M = (x^2 – 2xy + y^2) + (4x^2 – 4x +1 ) + 2022

   M = (x^2 – 2 .x.y + y^2) + [(2x)^2 – 2. 2x.1 + 1^2] + 2022

   M = (x-y)^2 + (2x-1)^2 + 2022

   Vì $\{\begin{array}{l}(x-y{)}^2\ge 0\mathrm{\forall }x,y\in \mathbb{R}\\ (2x-1{)}^2\ge 0\mathrm{\forall }x\in \mathbb{R}\end{array}$

   => (x-y)^2 + (2x-1)^2 >= 0\ AA x in RR

   => (x-y)^2 + (2x-1)^2 + 2022 >= 2022\ AA x in RR

   Dấu $”=”$ xảy ra khi $\{\begin{array}{l}(x-y{)}^2=0\\ (2x-1{)}^2=0\end{array}$

   <=> $\{\begin{array}{l}x-y=0\\ 2x-1=0\end{array}$

   <=> $\{\begin{array}{l}y={\displaystyle \frac{1}{2}}\\ x={\displaystyle \frac{1}{2}}\end{array}$

   Vậy $M_{min}=2022$ khi $x=y={\displaystyle \frac{1}{2}}$

   Trả lời
2. M=4x^2+y^2-4x-2xy+2023

   =x^2-2xy+y^2+4x^2-4x+1+2022

   =(x^2-2xy+y^2)+(4x^2-4x+1)+2022

   =(x-y)^2+(2x-1)^2+2022

   Vì (x-y)^2+(2x-1)^2>=0AA x,y in RR

   =>(x-y)^2+(2x-1)^2+2022>=0+2022AAx,y in RR

   =>A>=2022AAx,yinRR

   Dấu “=” xảy ra khi: {(x-y=0),(2x-1=0):}

   <=>{(x=y),(x=1/2):}

   <=>{(x=1/2),(y=1/2):}

   Vậy GTNN của biểu thức là 2022 khi {(x=1/2),(y=1/2):}

   Trả lời