(x+1)(x+3)(x+5)(x+7)+15=0高次方程.求详解以及检验后的最终答案!

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(x+1)(x+3)(x+5)(x+7)+15=0高次方程.求详解以及检验后的最终答案!(x+1)(x+3)(x+5)(x+7)+15=0高次方程.求详解以及检验后的最终答案!(x+1)(x+3)(x

(x+1)(x+3)(x+5)(x+7)+15=0高次方程.求详解以及检验后的最终答案!
(x+1)(x+3)(x+5)(x+7)+15=0
高次方程.求详解以及检验后的最终答案!

(x+1)(x+3)(x+5)(x+7)+15=0高次方程.求详解以及检验后的最终答案!
(x+3)(x+5)=x^2+8x+15
(x+1)(x+7)=x^2+8x+7
设(x+3)(x+5)=y
则:(x+1)(x+7)=y-8
原方程化为 y(y-8)+15=0
y^2-8y+15=0
(y-3)(y-5)=0
y=3或y=5
所以:(x+3)(x+5)=3或(x+3)(x+5)=5
当:(x+3)(x+5)=3时
x^2+8x+15=3
x^2+8x+12=0
(x+2)(x+6)=0
解得:x=-2或x=-6
当(x+3)(x+5)=5时
x^2+8x+15=5
x^2+8x+10=0
根据求根公式得:
x=根号6 -4 或 -根号6 -4
将上面解出的四个x的值代入原方程后,全部满足条件

(x+3)(x+5)=x^2+8x+15
(x+1)(x+7)=x^2+8x+7
设(x+3)(x+5)=y
则:(x+1)(x+7)=y-8
原方程化为 y(y-8)+15=0
y^2-8y+15=0
(y-3)(y-5)=0
y=3或y=5

(x+1)(x+3)(x+5)(x+7)+15=0,
(x^2+8x+7)(x^2+8x+15)+15=0,
令x^2+8x+11=t,方程成为
(t-4)(t+4)+15=0,
t^2=1,
t=±1.
(1).t=-1时,
x^2+8x+11=-1,x^2+8x+12=0,
x1=-2,x2=-6,
(2).t=1时,
x^2+8x+11=1,x^2+8x+10=0,
x3=-4+√6,x4=-4-√6.

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