Tìm x,y,z ta có:
(3x-5)^2018+(y^2-1)^2020+(5/3-z)^2022=0
Cho ctlhn+ vote 5 sao
Tìm x,y,z ta có: (3x-5)^2018+(y^2-1)^2020+(5/3-z)^2022=0 Cho ctlhn+ vote 5 sao
By Kylie
By Kylie
Tìm x,y,z ta có:
(3x-5)^2018+(y^2-1)^2020+(5/3-z)^2022=0
Cho ctlhn+ vote 5 sao
Giải thích các bước giải:
Ta có:
`(3x-5)^2018ge0` với mọi `x`
`(y^2-1)^2020ge0` với mọi `y`
`(5/3-z)^2020ge0` với mọi `z`
Mà `(3x-5)^2018+(y^2-1)^2020+(5/3-z)^2022=0`
`=>(3x-5)^2018=0`
`(y^2-1)^2020=0`
`(5/3-z)^2022=0`
`=>3x-5=0`
`y^2-1=0`
`5/3-z=0`
`=>3x=5`
`y^2=1`
`z5/3`
`=>x=5/3`
`y=+-1`
`z=5/3`
Có:
(3x-5)^2018≥0
(y^2-1)^2020≥0
(5/3-z)^2022≥0
⇒(3x-5)^2018+(y^2-1)^2020+(5/3-z)^2022 ≥0
Dấu “=” xảy ra khi: (3x-5)^2018 =0 ⇒3x-5=0 ⇒x=5/3
(y^2-1)^2020=0 ⇒y^2-1=0⇒y^2=1^2⇒y=±1
(5/3-z)^2022=0 ⇒5/3-z=0 ⇒z=5/3
Nhớ vote 5* và ctlhn nhé!