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222-9+11+12:2*14+14 = ? ( )

Toán Lớp 9: phương trình căn x- 2020 + y – 2021 + z- 2022 = 1/2( x +y + z ) -3030 có nghiệm là

Toán Lớp 9: phương trình căn x- 2020 + y – 2021 + z- 2022 = 1/2( x +y + z ) -3030 có nghiệm là

Comments ( 2 )

  1. \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):}

  2. Giải đáp + Lời giải và giải thích chi tiết:
    \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):}

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222-9+11+12:2*14+14 = ? ( )

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