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

Toán Lớp 8: B1 :phân tích đa thức thành nhân tử a) (9x^2 +1) – 36x^2 b) 9x^2 +6xy – 8y^2 c) x^3 – 9x^2 + 27x – 28 d)2x^3 – 9x^2 + 27x – 27 B2: vs m

Toán Lớp 8: B1 :phân tích đa thức thành nhân tử
a) (9x^2 +1) – 36x^2
b) 9x^2 +6xy – 8y^2
c) x^3 – 9x^2 + 27x – 28
d)2x^3 – 9x^2 + 27x – 27
B2: vs mọi nguyên n CMR:
a) A= 9-(2+5n)^2 chia hết cho 5
b) B= (2n-3)^2 – 25

Comments ( 2 )

  1. a)(9x^2+1)^2-36x^2
    =(9x^2-6x+1)(9x^2+6x+1)
    =(3x-1)^2(3x+1)^2
    b)9x^2+6xy-8y^2
    =(3x-2y)(4y+3x)
    c)x^3-9x^2+27x-28
    =(x-3)^3-1
    =(x-3-1)(x^2-6x+9+x-3+1)
    =(x-4)(x^2-5x+7)
    d)2x^3-9x^2+27x-27
    =x^3+x^3-9x^2+27x-27
    =x^3+(x-3)^3
    =(x+x-3)(x^2-x^2+3x+x^2-6x+9)
    =(2x-3)(x^2-3x+9)
    Bài 2:
    a)A=9-(2+5n)^2
    =9-4+20n+25n^2
    =5+20n+25n^2
    =5(1+4n+5n^2)\vdots5(đpcm)
    b)B=(2n-3)^2-25
    =(2n-3-5)(2n-3+5)
    =(2n-8)(2n+2)
    ->Không rõ đề

  2. Giải đáp:
     Bài 1
    a, (9x^2 + 1)^2 – 36x^2
    = (9x^2 + 1)^2 – (6x)^2
    = (9x^2 + 1 + 6x)(9x^2 + 1 – 6x)
    = [(3x)^2 + 2.x.1 + 1^2][(3x)^2 – 2.3x.1 + 1^2]
    = (3x + 1)^2 (3x – 1)^2
    b, 9x^2 + 6xy – 8y^2
    = 9x^2 + 6xy + y^2 – 9y^2
    = [(3x)^2 + 2.3x.y + y^2] – 9y^2
    = (3x + y)^2 – (3y)^2
    = (3x + 3y + y)(3x + y – 3y)
    = (3x + 4y)(3x – 2y)
    c, x^3 – 9x^2 + 27x – 27
    = x^3 – 3.x^2 .3 + 3.x.3^2 – 3^3
    = (x – 3)^3
    d, 2x^3 – 9x^2 + 27x – 27
    = x^3 + (x^3 – 9x^2 + 27x – 27)
    = x^3 + (x – 3)^2
    = (x + x – 3)[x^2 – x(x – 3) + (x – 3)^2]
    = (2x – 3)(x^2 – x^2 + 3x + x^2 – 6x + 9)
    = (2x – 3)(x^2 – 3x + 9)
    Bài 2:
    a, A = 9 – (2 + 5n)^2
    = 9 – [2^2 + 2.2.5n + (5n)^2]
    = 9 – (4 + 20n + 25n^2)
    = 9 – 4 – 20n – 25n^2
    = 5 – 20n – 25n^2
    = 5(1 – 4n – 5n^2)$\vdots$ 5
    → đpcm
    Vậy A = 9 – (2 + 5n)^2 $\vdots$ 5
    b, B = (2n – 3)^2 – 25
    =(2n)^2 – 2.2n.3 + 3^2 – 25
    = 4n^2 – 12n + 9 – 25
    = 4n^2 – 12n – 16
    → Thiếu đề

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