(x+1)(x+2)(x+3)-6*7*8用换元法分解因式
来源:学生作业帮 编辑:大师作文网作业帮 分类:数学作业 时间:2024/09/24 22:38:56
(x+1)(x+2)(x+3)-6*7*8用换元法分解因式
/>令x+2=t,则x+1=t-1,x+3=t+1
(x+1)(x+2)(x+3)-6×7×8
=(t-1)t(t+1)-(7-1)×7×(7+1)
=t(t^2-1)-7×(7^2-1)
=t^3-t-7^3+7
=(t^3-7^3)-(t-7)
=(t-7)(t^2+7t+49)-(t-7)
=(t-7)(t^2+7t+48)
=(x+2-7)[(x+2)^2+7(x+2)+48]
=(x-5)(x^2+4x+4+7x+14+48)
=(x-5)(x^2+11x+66)
(x+1)(x+2)(x+3)-6×7×8
=(t-1)t(t+1)-(7-1)×7×(7+1)
=t(t^2-1)-7×(7^2-1)
=t^3-t-7^3+7
=(t^3-7^3)-(t-7)
=(t-7)(t^2+7t+49)-(t-7)
=(t-7)(t^2+7t+48)
=(x+2-7)[(x+2)^2+7(x+2)+48]
=(x-5)(x^2+4x+4+7x+14+48)
=(x-5)(x^2+11x+66)
1;分解因式;(x+1)(x+2)(X+3)-6*7*8
分解因式 (1)x^2+y^2+6x-2y+8 (2) (x^2+3x+3)(x^2+3x+7
分解因式(x^2+4x)^2-8(x^2+4x+6)
分解因式:(x+1)(x+2)(x+3)(x+4)+1.
分解因式 (x²+3x+2)*(x²+7x+12)—120
分解因式(x + 1)四次方 + (x² + 4x + 8)+ 2x²
分解因式:2x^4-x^3-6x^2-x+2(换元法)
若把x²+3x+c分解因式得到(x+1)(x+2),则多项式x²-3x+c分解因式为
分解因式:(x+1)(x+2)(x+3)(x+6)+x2=______.
分解因式:8(x^2-2y^2)-x(7x+y)+xy=
分解因式:8(x²-2y²)-x(7x+y)+xy
分解因式8(x²-2y²)-x(7x+y)+xy=