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

Question

bichha · Answer

`(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)`

Trả lời

lanhoa · Answer

Đ&aacute;p &aacute;n:   `(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)`

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

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