`x^2+|x-2|+4y^2-4xy=0` giải rõ

Ta có: `x^2 + |x - 2| + 4y^2 - 4xy = 0`

`<=> (x^2 - 4xy + 4y^2) + |x - 2| = 0`

`<=> (x - 2y)^2 + |x - 2| = 0`

Vì `(x - 2y)^2 \ge 0 AAx,y` 

    `|x - 2| \ge 0 AAx`

`=> (x - 2y)^2 + |x - 2| \ge 0`

Dấu "`=`" xảy ra `<=> {((x - 2y)^2 = 0),(|x - 2| = 0):}`

`<=> {(x - 2y = 0),(x - 2=  0):}`

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

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

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

Vậy `(x,y) = (2,1)`

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