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

Toán Lớp 8: Cho $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$. CMR: $\frac{1}{x^{2021}}+\frac{1}{y^{2021}}+\frac{1}{z^{2021}}=\frac{1}{x^

Tháng Ba 18, 2022 Toán học 1 Comment 0 views

Toán Lớp 8: Cho $\frac{1}{x}+\frac{1}{y}+\frac{1}{z}=\frac{1}{x+y+z}$. CMR: $\frac{1}{x^{2021}}+\frac{1}{y^{2021}}+\frac{1}{z^{2021}}=\frac{1}{x^{2021}+y^{2021}+z^{2021}}$

Comments ( 1 )

  1. ngoclandiep
    ngoclandiep
    18 Tháng Ba, 2022 at 11:00 sáng
    Reply
    $\\$
    1/x+1/y+1/z=1/(x+y+z)
    <=> (xy+yz+xz)/(xyz) – 1/(x+y+z)=0
    <=> ((xy+yz+xz)(x+y+z))/(xyz(x+y+z))-(xyz)/(xyz(x+y+z))=0
    <=> x^2y + xyz +x^2z + xy^2 + y^2z +xyz + xyz + yz^2 + xz^2 – xyz =0
    <=>x^2y + x^2z + xy^2 + y^2z + yz^2 +xz^2 +2xyz=0
    <=>x^2 (y+z) + (xy^2+xyz) + yz (y+z) + (xz^2+xyz)=0
    <=>x^2(y+z) + xy (y+z) + yz (y+z) + xz (y+z)=0
    <=> (y+z)(x^2 +xy+yz+xz)=0
    <=>(y+z)[x(x+y)+z(x+y)]=0
    <=> (x+y)(x+z) (y+z)=0
    Không mất tính tổng quát giả sử x+y=0
    <=>x=-y
    1/x^{2021}+1/y^{2021}+1/z^{2021}
    =-1/y^{2021}+1/y^{2021}+1/z^{2021}
    =1/z^{2021}(1)
    1/(x^{2021}+y^{2021}+z^{2021})
    =1/(-y^{2021}+y^{2021}+z^{2021})
    =1/z^{2021}(2)
    (1)(2)
    =>1/x^{2021}+1/y^{2021}+1/z^{2021}=1/(x^{2021}+y^{2021}+z^{2021})
     

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

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