Question

Answer 1

`a^2 + b^2 - 2a + 4b + 5 = 0`

`<=> (a^2 - 2a + 1) +(b^2 + 4b  + 4) = 0`

`<=> (a - 1)^2 + (b+2)^2 = 0`

Do `{((a - 1)^2 >=0),((b+2)^2 >= 0):}`

`=>  (a - 1)^2 + (b+2)^2 >= 0`

Dấu = có khi: `{((a - 1)^2 =0),((b+2)^2 = 0):}`

` <=> {(a - 1 =0),(b+2= 0):}`

`<=>  {(a=1),(b=-2):}`

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