Cho `x^2+y^2+z^2=4` và `x^3+y^3+z^3=8` hãy tính `x^4+y^4+z^4` 25/11/2021 Bởi Iris Cho `x^2+y^2+z^2=4` và `x^3+y^3+z^3=8` hãy tính `x^4+y^4+z^4`
`2(x^2 + y^2 + z^2) = x^3 + y^3 + z^3``⇒ 2x^2 + 2y^2 + 2z^2 = x^3 + y^3 + z^3``⇒ x^3 − 2x^2 + y^3 − 2y^2 + z^3 − 2z^2 = 0``⇒ x^2(x − 2) + y^2(y − 2) + z^2(z − 2) = 0` mặt khác `:x^2 + y^2 + z^2 = 4` `⇒ x^2 ≤ 4, y^2 ≤ 4, z^2 ≤ 4``⇒ -2≤x ≤ 2, -2≤y ≤ 2, -2≤z ≤ 2``⇒ x − 2 ≤ 0, y − 2 ≤ 0, z − 2 ≤ 0``⇒ x^2(x − 2) ≤ 0, y^2(y − 2) ≤ 0, z^2(z − 2) ≤ 0``⇒ x^2(x − 2) = 0 ,y^2(y − 2) = 0 ,z^2(z − 2) = 0``⇒ (z; y; z) = (2; 0; 0)= (0; 0; 2)= (0; 2; 0)``⇒ x^4 + y^4 + z^4 = 16` Bình luận
`2(x^2 + y^2 + z^2) = x^3 + y^3 + z^3`
`⇒ 2x^2 + 2y^2 + 2z^2 = x^3 + y^3 + z^3`
`⇒ x^3 − 2x^2 + y^3 − 2y^2 + z^3 − 2z^2 = 0`
`⇒ x^2(x − 2) + y^2(y − 2) + z^2(z − 2) = 0`
mặt khác `:x^2 + y^2 + z^2 = 4`
`⇒ x^2 ≤ 4, y^2 ≤ 4, z^2 ≤ 4`
`⇒ -2≤x ≤ 2, -2≤y ≤ 2, -2≤z ≤ 2`
`⇒ x − 2 ≤ 0, y − 2 ≤ 0, z − 2 ≤ 0`
`⇒ x^2(x − 2) ≤ 0, y^2(y − 2) ≤ 0, z^2(z − 2) ≤ 0`
`⇒ x^2(x − 2) = 0 ,y^2(y − 2) = 0 ,z^2(z − 2) = 0`
`⇒ (z; y; z) = (2; 0; 0)= (0; 0; 2)= (0; 2; 0)`
`⇒ x^4 + y^4 + z^4 = 16`