a) Ta có: \(\left(x-6\right)^2+207\ge207\left(\forall x\right)\)
Dấu "=" xảy ra khi: \(\left(x-6\right)^2=0\Rightarrow x=6\)
Vậy Min = 207 khi x = 6
b) Ta có: \(\left(x+5\right)^2+\left(y-9\right)^2+2021\ge2021\left(\forall x,y\right)\)
Dấu "=" xảy ra khi: \(\hept{\begin{cases}\left(x+5\right)^2=0\\\left(y-9\right)^2=0\end{cases}}\Rightarrow\hept{\begin{cases}x=-5\\y=9\end{cases}}\)
Vậy Min = 2021 khi \(\hept{\begin{cases}x=-5\\y=9\end{cases}}\)
Trả lời :
a, A = (x - 6)2 + 207
Do (x - 6)2 > 0 => (x - 6)2 + 207 > 207
Dấu "=" xảy ra <=> (x - 6)2 + 207 = 207
<=> (x - 6)2 = 0 <=> x = 6.
Vậy MinA = 207 <=> x = 6.
