`a)`
Ta có :
`(a-b)^2 + 4ab `
` = a^2 - 2ab + b^2 + 4ab`
` = a^2 + b^2 + 2ab`
` = (a+b)^2`
Vậy `(a+b)^2 = (a-b)^2 + 4ab`
`b)`
Ta có :
`(a+b)^2 - 4ab`
` = a^2 + 2ab + b^2 - 4ab`
` = a^2 + b^2 - 2ab`
` = (a-b)^2`
Vậy `(a-b)^2 = (a+b)^2 - 4ab`
`c)`
Ta có :
`(ax - by)^2 +(ay + bx)^2 `
` = (ax)^2 - 2axby + (by)^2 + (ay)^2 + 2aybx + (bx)^2`
` = a^2x^2 - 2axby + b^2y^2 + a^2y^2 +2axby + b^2x^2`
` = a^2x^2 + b^2y^2 + a^2y^2+ b^2x^2`
` = ( a^2x^2 + b^2x^2) + (b^2y^2 + a^2y^2)`
` = x^2 . (a^2 + b^2) + y^2 . (b^2 + a^2)`
` = (a^2 +b^2) . (x^2 +y^2)`
Vậy `(a^2 +b^2) . (x^2 +y^2) = (ax - by)^2 +(ay + bx)^2`
`d)`
Ta có :
`a^2 +b^2 + c^2 + 3 = 2(a+b+c)`
`=> a^2 + b^2 + c^2 + 3 = 2a + 2b + 2c`
`=> a^2 + b^2 +c^2 + 3 - 2a - 2b - 2c =0`
=> (a^2 - 2a + 1) + (b^2 - 2b + 1) + (c^2 - 2c + 1) =0`
`=> (a-1)^2 + (b-1)^2 + (c-1)^2 = 0`
`\forall a ; b ; c` ta có :
`(a-1)^2 \ge 0`
`(b-1)^2 \ge 0`
`(c-1)^2 \ge 0`
`=> (a-1)^2 + (b-1)^2 + (c-1)^2 \ge 0`
Dấu `=` xảy ra `<=>` $\left\{\begin{matrix} a - 1 = 0 \\ b-1=0 \\ c-1=0 \end{matrix}\right.$
`<=>` $\left\{\begin{matrix} a = 1 \\ b =1 \\ c = 1\end{matrix}\right.$
Vậy nếu `a^2 +b^2 + c^2 + 3 = 2(a+b+c)` thì `a = b =c =1`
