Question

a^2 +16a+b^2+6b+73

Answer 1

`a^2 + 16a + b^2 + 6b + 73`

`= a^2 + 16a + 64 + b^2 + 6b + 9`

`= (a^2 + 16a + 64) + (b^2 + 6b +9)`

`= (a+8)^2 + (b+3)^2`

Do `{( (a +8)^2>=0),((b + 3)^2  >= 0):}`

`=>  (a +8)^2 + (b + 3)^2 >= 0`

Dấu = có khi: 

`{( a +8=0),(b + 3= 0):}`

`{( a =-80),(b =-3):}`

Vậy ...

Answer 2

\[
(a^2 + 16a) + (b^2 + 6b) + 73
\]
\[
a^2 + 16a = (a + 8)^2 - 64
\]
\[
b^2 + 6b = (b + 3)^2 - 9
\]
\[
(a + 8)^2 - 64 + (b + 3)^2 - 9 + 73
\]
\[
-64 - 9 + 73 = 0
\]
\[
(a + 8)^2 + (b + 3)^2
\]
\[
(a + 8)^2 + (b + 3)^2
\]
=> a = -8 ; b = -3