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

Toán Lớp 8: Bài 7: C/m rằng với mọi số nguyên x, y thì: a) A= x ( $x^{2}$ +x) +x (x+1) chia hết cho x+1. b) B= $xy^{2}$ – $yx^{2}$ +xy chia hế

Toán Lớp 8: Bài 7: C/m rằng với mọi số nguyên x, y thì:
a) A= x ( $x^{2}$ +x) +x (x+1) chia hết cho x+1.
b) B= $xy^{2}$ – $yx^{2}$ +xy chia hết cho xy.
c) C= $x^{2}$ $y^{3}$ + $x^{3}$ $y^{3}$ – $xy^{2}$ chia hết cho $x^{2}$y +xy -1

Comments ( 2 )

  1. Giải đáp:
     Bài 7:
    a, A = x(x^2 + x) + x(x + 1)
    = x^2 (x + 1) + x(x + 1)
    = (x + 1)(x^2 + x)
    = (x + 1)(x + 1)x
    = (x + 1)^2 x
    Mà (x + 1)^2 x $\vdots$ (x + 1)
    Vậy A $\vdots$ (x + 1) (đpcm)
    b, B = xy^2 – yx^2 + xy
    = xy (y – x + 1)
    Mà xy(y – x + 1) $\vdots$ xy
    Vậy B $\vdots$ xy (đpcm)
    c, C = x^2 y^3 + x^3 y^3 – xy^2
    = xy^2 .xy + xy^2 . x^2 y – xy^2 . 1
    = xy^2 (xy + x^2 y – 1)
    Mà xy^2 (xy + x^2 y – 1) $\vdots$ (x^2 y + xy – 1)
    Vậy C $\vdots$ (x^2 y + xy – 1) (đpcm)

  2. a)
    A=x(x^2+x)+x(x+1)
    =x^2(x+1)+x(x+1)
    =(x^2+x)(x+1)
    =x(x+1)(x+1)
    =x(x+1)^2 \ vdots \ x+1
    to đpcm
    b)
    B=xy^2-yx^2+xy
    =xy.y-xy.x+xy
    =xy(y-x+1) \ vdots \ xy
    to đpcm
    c)
    C=x^2y^3+x^3y^3-xy^2
    =xy^2 . xy + xy^2 . x^2y – xy^2
    =xy^2 (x^2y + xy – 1) \ vdots \ x^2y+xy-1
    to đpcm

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