Question

[Minh Thắng Jr.]

Phân tích các đa thức sau thành nhân tử:

P (x) = x(x + 2)(x + 3)(x + 5) - 7

Q(x) = (4x -2)(10x + 4)(5x + 7)(2x + 1) + 17

R (x) = (3x + 2)(3x -5)(x - 1)(9x + 10) + 24x²

H (x) = (x - 18)(x - 7)(x + 35)(x + 90) - 67x²

K (x) = (6x + 5)²(3x + 2)(x + 1) - 35

G (x) = (4x + 1)(12x -1)(3x + 2)(x + 1) - 4

I (x) = (x² + 5x + 6)(x²- 15x + 56) -144

M (x) = (x²- 11x + 28)(x^{2 - 7x + 10) - 72}

Mong mọi người gúp đỡ hộ bài này với ạ Cảm ơn mọi người

Answer 1

[Akai Haruma]

Lời giải:

\(P(x)=x(x+2)(x+3)(x+5)-7\)

\(=[x(x+5)][(x+2)(x+3)]-7\)

\(=(x^2+5x)(x^2+5x+6)-7\)

\(=a(a+6)-7\) (đặt \(x^2+5x=a\) )

\(=a^2+6a-7=a^2-a+7a-7\)

\(=a(a-1)+7(a-1)=(a-1)(a+7)\)

\(=(x^2+5x-1)(x^2+5x+7)\)

-----------------

\(Q(x)=(4x-2)(10x+4)(5x+7)(2x+1)+17\)

\(=4(2x-1)(5x+2)(5x+7)(2x+1)+17\)

\(=4[(2x-1)(5x+7)][(5x+2)(2x+1)]+17\)

\(=4(10x^2+9x-7)(10x^2+9x+2)+17\)

\(=4a(a+9)+17\) (đặt \(10x^2+9x-7=a\)

\(=4a^2+36a+17=(2a+9)^2-8^2\)

\(=(2a+9-8)(2a+9+8)=(2a+1)(2a+17)\)

\(=(20x^2+18x-13)(20x^2+18x+3)\)

Answer 2

[Akai Haruma]

\(R(x)=(3x+2)(3x-5)(x-1)(9x+10)+24x^2\)

\(=[(3x+2)(3x-5)][(x-1)(9x+10)]+24x^2\)

\(=(9x^2-9x-10)(9x^2+x-10)+24x^2\)

\(=(a-9x)(a+x)+24x^2\) (đặt \(9x^2-10=a\) )

\(=a^2-8ax+15x^2=(a^2-5ax)-(3ax-15x^2)\)

\(=a(a-5x)-3x(a-5x)=(a-3x)(a-5x)\)

\(=(9x^2-3x-10)(9x^2-5x-10)\)

--------------------------

\(H(x)=(x-18)(x-7)(x+35)(x+90)-67x^2\)

\(=[(x-18)(x+35)][(x-7)(x+90)]-67x^2\)

\(=(x^2+17x-630)(x^2+83x-630)-67x^2\)

\(=a(a+66x)-67x^2\) (đặt \(x^2+17x-630=a\) )

\(=a^2-ax+67ax-67x^2\)

\(=a(a-x)+67x(a-x)=(a-x)(a+67x)\)

\(=(x^2+16x-630)(x^2+84x-630)\)