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

Toán Lớp 8: Cho a+b+c=0; CMR: a ³+b ³+c ³=3abc

Toán Lớp 8: Cho a+b+c=0; CMR: a ³+b ³+c ³=3abc

Comments ( 2 )

  1. Giải đáp:
     Ta có:
    Ta có:
    a + b + c = 0
     ⇒ a + b = -c
    Từ đó suy ra:
    a^3 + b^3 + c^3
    = (a + b)^3 + c^3 -3ab(a + b)
    =  (-c)^3 + c^3 – 3ab(a+b)
    = -3ab . -c
    = 3abc (đpcm)

  2. Cách 1:
    a³ + b³ + c³ – 3abc = a³ + 3ab(a + b) + b³ + c³ – 3abc – 3ab(a + b)
    = (a + b)³ + c³ – 3ab(a + b + c)
    = (a + b + c)(a² + 2ab + b² – ca – bc + c²) – 3ab(a + b + c)
    = (a + b + c)(a² + b² + c² – ab – bc – ca)
    Vậy a³ + b³ + c³ – 3abc = (a + b + c)(a² + b² + c² – ab – bc – ca)
    Cách 2:

    a³ + b³ + c³ – 3abc = ( a³ + b³ ) + c³ – 3abc
    = ( a + b)³ – 3ab( a + b) + c³ – 3abc
    = ( a + b)³ – 3a²b – 3ab² + c³ – 3abc
    = {(a + b)³ + c³ } -3a²b – 3ab² – 3abc
    = ( a + b + c)³ – 3c(a + b)( a + b + c ) – 3ab(a + b+ c)
    = (a + b + c) {( a + b + c)² -3c(a +b) – 3ab}
    = (a + b + c) {a² + b² + c² +2ab +2bc+ 2ca -3ca – 3bc – 3ab}
    = (a + b + c)( a² + b² + c² – ab – bc – ca)
     
     

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

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