Cho P = 3/x-3 - 6x/9-x^2 + x/x+3 a.Rút gọn P b.Tìm x để P = 0

`a)` `P=3/{x-3} - {6x}/{9-x^2} + x/{x+3}` `(x \ne ± 3)`

`=3/{x-3} - {-6x}/{9-x^2} + x/{x+3}`

`= {3(x+3)}/{(x-3)(x+3)} - {-6x}/{(x-3)(x+3)} + {x(x-3)}/{(x-3)(x+3)}`

`= {3x+9}/{(x-3)(x+3)} - {-6x}/{(x-3)(x+3)} + {x^2-3x}/{(x-3)(x+3)}`

`= {3x+9+6x+x^2-3x}/{(x-3)(x+3)}`

`= {x^2+6x+9}/{(x-3)(x+3)}`

`= {(x+3)^2}/{(x-3)(x+3)}`

`= {(x+3)}/{(x-3)}`

`b)` Để `P=0` 

`<=> {(x+3)}/{(x-3)} =0`

`<=> x+3=0` (vì `x-3 > 0 ∀ x`)

`<=> x=-3` (không thoả mãn)

Vậy không có giá trị của `x` để `P=0`

Answer

`a, P = 3/{x - 3} - {6x}/{9 - x^2} + x/{x + 3}` `(Đk: x \ne +-3)`

`=> P = 3/{x - 3} + {6x}/{x^2 - 9} + x/{x + 3}`

`=> P = 3/{x - 3} + {6x}/{x^2 - 3^2} + x/{x + 3}`

`=> P = 3/{x - 3} + {6x}/{(x - 3) . (x + 3)} + x/{x + 3}`

`=> P = {3 . (x + 3)}/{(x - 3) . (x + 3)} + {6x}/{(x - 3) . (x + 3)} + {x . (x - 3)}/{(x - 3) . (x + 3)}`

`=> P = {3x + 9}/{(x - 3) . (x + 3)} + {6x}/{(x - 3) . (x + 3)} + {x^2 - 3x}/{(x - 3) . (x + 3)}`

`=> P = {3x + 9 + 6x + x^2 - 3x}/{(x - 3) . (x + 3)}`

`=> P = {x^2 + (3x + 6x - 3x) + 9}/{(x - 3) . (x + 3)}`

`=> P = {x^2 + 6x + 9}/{(x - 3) . (x + 3)}`

`=> P = {x^2 + 2 . x . 3 + 3^2}/{(x - 3) . (x + 3)}`

`=> P = {(x + 3)^2}/{(x - 3) . (x + 3)}`

`=> P = {x + 3}/{x - 3}`

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`b,` Tìm `x` để `P = 0`

`=> 0 = {x + 3}/{x - 3}`

`=> x + 3 = 0`

`=> x = 0 - 3`

`=> x = -3` (KTM)

Vậy `x \in` $\varnothing$