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`1.` ` x^2 + y^2 + 4y + 13 = 6x `
` (x^2 - 6x +9 ) ( y^2 + 4y 4 ) = 0 `
` (x - 3)^2 + (y + 2)^2 = 0 `
`{( (x -3)^2 = 0),( (y + 2)^2 = 0):}` `{( x - 3 = 0),( y + 2 = 0):}` `{(x = 3),(y = -2):}`
`=>` `x = 3` , `y = -2`
`2.` ` x^2 + y^2 + 17 = 2x - 8y `
` (x^2 - 2x +1) + (y^2 + 8y + 16) = 0 `
` (x - 1)^2 + (y + 4)^2 = 0 `
Vì `(x - 1)^2 >= 0` ; `(y + 4)^2 >= 0`
Để `(x - 1)^2 + (y + 4)^2 = 0` thì `:`
`{(x - 1 = 0),(y + 4 = 0):}` `=>` `{(x = 1),(y = -4):}`
Vậy `x = 1` ; `y = -4`
`3.` `x^2 + y^2 + 45 = 12y - 6x`
`( x^2 + 6x + 9 ) + ( y^2 - 12y + 36 ) = 0`
`( x + 3 )^2 + (y - 6)^2 = 0
Vì `(x + 3)^2 >= 0` ; `(y - 6)^2 >= 0`
Để `(x + 3)^2 + (y - 6 )^2 >= 0` thì `:`
`{(x + 3 = 0),(y - 6 = 0):}` `=>` `{(x = -3),(y = 6):}`
Vậy `x = -3` ; `y = 6`
`4.` `4x^2 + 9y^2 + 2 = 4x + 6y`
`(4x^2 - 4x +1) + (9y^2 -6y +1) = 0`
`(2x - 1)^2 + (3y - 1)^2 = 0`
Vì `(2x - 1)^2 >= 0` ; `(3y - 1)^2 >= 0`
Để `(2x - 1)^2 + (3y - 1)^2 = 0` thì `:`
`{(2x - 1 = 0),(3y - 1 = 0):}` `=>` `{(x = 1/2),(y = 1/3):}`
Vậy `x = 1/2` ; `y = 1/3`
`5.` `9x^2 + 4y^2 + 26 + 4y = 30x`
`(9x^2 - 30x + 25) + (4y^2 + 4y + 1) = 0`
`(3x - 5)^2 + (2y + 1)^2 = 0`
`=>` `{( (3x - 5)^2 = 0),( (2y +1)^2 = 0):}`
`<=>` `{(3x - 5 = 0),(2y + 1 = 0):}` `<=>` `{(x = 5/3),(y = -1/2):}`
Vậy `x = 5/3` ; `y = -1/2`
`6.` `9x^2 + y^2 + 20 = 12x + 8y`
`(9x^2 - 12x + 4) + (y^2 - 8y +16) = 0`
`(3x - 2)^2) + (y - 4)^2 = 0`
`=>` `{( (3x - 2)^2 = 0),( (y - 4)^2 = 0):}` `=>` `{(x = 2/3),(y = 4):}`
Vậy `x = 2/3` ; `y = 4`
Hãy giúp mọi người biết câu trả lời này thế nào?
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1) ` x^2 + y^2 + 4y + 13 = 6x `
` (x^2 - 6x +9 ) ( y^2 + 4y 4 ) = 0 `
` (x - 3)^2 + (y + 2)^2 = 0 `
`{( (x -3)^2 = 0),( (y + 2)^2 = 0):}` `{( x - 3 = 0),( y + 2 = 0):}` `{(x = 3),(y = -2):}`
`->` vậy `x = 3` , `y = -2`
2) ` x^2 + y^2 + 17 = 2x - 8y `
` (x^2 - 2x +1) + (y^2 + 8y + 16) = 0 `
` (x - 1)^2 + (y + 4)^2 = 0 `
Vì `(x - 1)^2 >= 0` ; `(y + 4)^2 >= 0`
Để `(x - 1)^2 + (y + 4)^2 = 0` thì `:`
`{(x - 1 = 0),(y + 4 = 0):}` `=>` `{(x = 1),(y = -4):}`
`->` vậy `x = 1` ; `y = -4`
3) `x^2 + y^2 + 45 = 12y - 6x`
`( x^2 + 6x + 9 ) + ( y^2 - 12y + 36 ) = 0`
`( x + 3 )^2 + (y - 6)^2 = 0`
Vì `(x + 3)^2 >= 0` ; `(y - 6)^2 >= 0`
Để `(x + 3)^2 + (y - 6 )^2 >= 0` thì `:`
`{(x + 3 = 0),(y - 6 = 0):}` `=>` `{(x = -3),(y = 6):}`
`->` vậy `x = -3` ; `y = 6`
4) `4x^2 + 9y^2 + 2 = 4x + 6y`
`(4x^2 - 4x +1) + (9y^2 -6y +1) = 0`
`(2x - 1)^2 + (3y - 1)^2 = 0`
Vì `(2x - 1)^2 >= 0` ; `(3y - 1)^2 >= 0`
Để `(2x - 1)^2 + (3y - 1)^2 = 0` thì `:`
`{(2x - 1 = 0),(3y - 1 = 0):}` `=>` `{(x = 1/2),(y = 1/3):}`
`->` vậy `x = 1/2` ; `y = 1/3`
5) `9x^2 + 4y^2 + 26 + 4y = 30x`
`(9x^2 - 30x + 25) + (4y^2 + 4y + 1) = 0`
`(3x - 5)^2 + (2y + 1)^2 = 0`
`=>` `{( (3x - 5)^2 = 0),( (2y +1)^2 = 0):}`
`<=>` `{(3x - 5 = 0),(2y + 1 = 0):}` `<=>` `{(x = 5/3),(y = -1/2):}`
`->` vậy `x = 5/3` ; `y = -1/2`
6) `9x^2 + y^2 + 20 = 12x + 8y`
`(9x^2 - 12x + 4) + (y^2 - 8y +16) = 0`
`(3x - 2)^2) + (y - 4)^2 = 0`
`=>` `{( (3x - 2)^2 = 0),( (y - 4)^2 = 0):}` `=>` `{(x = 2/3),(y = 4):}`
`->` vậy `x = 2/3` ; `y = 4`
Hãy giúp mọi người biết câu trả lời này thế nào?
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