{x + y + z =9 {1/x + 1/y + 1/z=1 {xy + yz + xz = 27 Giai he phuong trinh

Đáp án:

$\left \{ {{\left \{ {{x + y + z = 9 } \atop {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1}} \right.} \atop {xy + yz + zx = 27}} \right.$  `↔` $\left \{ {{\left \{ {{(x + y + z)^2 = 81 (1) } \atop {\dfrac{1}{x} + \dfrac{1}{y} + \dfrac{1}{z} = 1 (2)}} \right.} \atop {3(xy + yz + zx) = 81 (3)}} \right.$  

Từ `(1)(3) -> (x + y + z)^2 = 3(xy + yz + zx)`

`↔ x^2 + y^2 + z^2 + 2(xy + yz + zx) – 3(xy + yz + zx) = 0`

`↔ x^2 + y^2 + z^2 – xy – yz – zx = 0`

`↔ 2x^2 + 2y^2 + 2z^2 – 2xy – 2yz – 2zx = 0`

`↔ (x^2 – 2xy + y^2) + (y^2 – 2yz + z^2) + (z^2 – 2zx + x^2) = 0`

`↔ (x – y)^2 + (y – z)^2 + (z – x)^2 = 0`

`↔` $\left \{ {{\left \{ {{x – y = 0 } \atop {y – z = 0}} \right.} \atop {z – x = 0}} \right.$  `↔ x = y = z (4)`

Từ `(1)(4) -> x = y = z = 3` ( thõa mãn `(2)`)

Vậy `x = y = z = 1`

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