phương trình căn x- 2020 + y - 2021 + z- 2022 = 1/2( x +y + z ) -3030 có nghiệm là

Đáp án + Giải thích các bước giải:

`\sqrt{x-2020}+\sqrt{y-2021}+\sqrt{z-2022}=1/2 (x+y+z)-3030`

`⇔2\sqrt{x-2020}+2\sqrt{y-2021}+2\sqrt{z-2022}= x+y+z-6060`

`⇔(x-2020-2\sqrt{x-2020}+1)+(y-2021-2\sqrt{y-2021}+1)+(z-2022-2\sqrt{x-2022}+1)=0`

`⇔(\sqrt{x-2020}-1)^2+(\sqrt{y-2021}-1)^2+(\sqrt{z-2022}-1)^2=0`

`⇔{(\sqrt{x-2020}-1=0),(\sqrt{y-2021}-1=0),(\sqrt{z-2022}-1=0):}`

`⇔{(\sqrt{x-2020}=1),(\sqrt{y-2021}=1),(\sqrt{z-2022}=1):}`

`⇔{(x-2020=1),(y-2021=1),(z-2022=1):}`

`⇔{(x=2021),(y=2022),(z=2023):}`

Vậy pt có nghiệm `{(x=2021),(y=2022),(z=2023):}`

`\sqrt{x - 2020 } + \sqrt{y - 2021} + \sqrt{z - 2022} = 1/2 ( x + y + z ) - 3030`

`⇔ 2\sqrt{x - 2020} + 2\sqrt{y - 2021} + 2\sqrt{z - 2022} = x + y + z - 6060`

`⇔ ( x - 2020 - 2\sqrt{x - 2020} + 1 ) + ( y - 2021 - 2\sqrt{y - 2021} + 1 ) + ( z - 2022 - 2 \sqrt{x - 2022} + 1 = 0`

`⇔  ( \sqrt{x - 2020} - 1^2 ) + ( \sqrt{y - 2021} -1 ) ^2 + ( \sqrt{z - 2022} - 1)^2 = 0`

`⇔ {(\sqrt{x - 2020} - 1 = 0),(\sqrt{y - 2021} - 1 = 0),(\sqrt{z - 2022} - 1 = 0):}`

`⇔ {(\sqrt{x - 2020} = 1 ),(\sqrt{y - 2021} =1),(\sqrt{z - 2022} = 1):}`

`⇔ {(x - 2020 = 1),(y - 2021 = 1),(z - 2022 = 1):}`

`⇔ {(x = 2021),(y = 2022),(z = 2023):}`

Vậy phương trình có nghiệm `{(x = 2021),(y = 2022),(z = 2023):}`