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

Toán Lớp 8: Cho x + y = 1. Tính $x^{3}$ + $y^{3}$ + 3xy

Toán Lớp 8: Cho x + y = 1. Tính $x^{3}$ + $y^{3}$ + 3xy

Comments ( 2 )

  1. Giải đáp:
     Ta có:
    x³ + y³ + 3xy
    = (x + y)^3 + 3xy(x + y) – 3xy (1)
    text{ Thay x + y = 1 vào (1), ta được:}
    1³ + 3xy . 1 – 3xy
    = 1 + 3xy – 3xy
    = 1 + (3xy – 3xy) = 1
    Vậy x³ + y³ + 3xy = 1 khi x + y = 1
     

  2. Theo đề bài:
    $x+y=1$<=>$x=1-y$
    Ta có:
    $x^{3}+y^{3}+3xy$
    $=(x+y)(x^{2}-xy+y^{2})+3xy$
    $=(1-y+y)[(1-y)^{2}-(1-y)y+y^{2}]+3(1-y)y$
    $=1(1-2y+y^{2}-y+y^{2}+y^{2})+3y-3y^{2}$
    $=1(1-3y+3y^{2})+3y-3y^{2}$
    $=1-3y+3y^{2}+3y-3y^{2}$
    $=1$
    Vậy $x+y=1$ thì  $x^{3}+y^{3}+3xy=1$

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

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