Pt đa thức thành nhân tử (x+2)(x+3)(x+4)(x+5)-24

0 bình luận về “Pt đa thức thành nhân tử (x+2)(x+3)(x+4)(x+5)-24”

1. `(x+2)(x+3)(x+4)(x+5)-24`

   `=[(x+2)(x+5)][(x+3)(x+4)]-24`

   `=(x^2+5x+2x+10)(x^2+4x+3x+12)-24`

   `=(x^2+7x+10)(x^2+7x+12)-24`

   `=(x^2+7x+11-1)(x^2+7x+11+1)-24`

   `=(x^2+7x+11)^2-1-24`

   `=(x^2+7x+11)^2-25`

   `=(x^2+7x+11)^2-5^2`

   `=(x^2+7x+11-5)(x^2+7x+11+5)`

   `=(x^2+7x+6)(x^2+7x+16)`

   `=(x^2+x+6x+6)(x^2+7x+16)`

   `=(x+1)(x+6)(x^2+7x+16)`

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2. Đáp án:

   `(x+2)(x+3)(x+4)(x+5)-24`

   `=[(x+2)(x+5)][(x+3)(x+4)]-24`

   `=(x^2+7x+10)(x^2+7x+12)-24`

   `=(x^2+7x+11)^2-1-24`

   `=(x^2+7x+11)^2-25`

   `=(x^2+7x+6)(x^2+7x+16)`

   `=(x^2+x+6x+6)(x^2+7x+16)`

   `=[x(x+1)+6(x+1)](x^2+7x+16)`

   `=(x+1)(x+6)(x^2+7x+16)`

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