(x+1)(x+2)(x+3)(x+4)=120这个方程怎么解?我正在学一元二次方程,帮帮忙.

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(x+1)(x+2)(x+3)(x+4)=120这个方程怎么解?我正在学一元二次方程,帮帮忙.(x+1)(x+2)(x+3)(x+4)=120这个方程怎么解?我正在学一元二次方程,帮帮忙.(x+1)(

(x+1)(x+2)(x+3)(x+4)=120这个方程怎么解?我正在学一元二次方程,帮帮忙.
(x+1)(x+2)(x+3)(x+4)=120这个方程怎么解?我正在学一元二次方程,帮帮忙.

(x+1)(x+2)(x+3)(x+4)=120这个方程怎么解?我正在学一元二次方程,帮帮忙.
(x+1)(x+2)(x+3)(x+4)=120 [ (x+1)(x+4)][(x+2)(x+3)]=120 (x^2+5x+4)(x^2+5x+6)=120
设(x^2+5x )=y (y+4)(y+6)=120 y^2+10y+24-120=0 y^2+10y-96=0 (y+16)(y-6)=0
y=-16 y=6 所以x^2+5x=-16或x^2+5x=6 x^2+5x=-16无解 x^2+5x=6
x^2+5x-6=0 x=-6 x=1

(x+1)(x+2)(x+3)(x+4)=120
(x²+5x+4)(x²+5x+6)=120
(x²+5x+4)(x²+5x+4+2)=120
(x²+5x+4)²+2(x²+5x+4)=120
(x²+5x+4)²+2(x²+5x+4)-120=0
(x&...

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(x+1)(x+2)(x+3)(x+4)=120
(x²+5x+4)(x²+5x+6)=120
(x²+5x+4)(x²+5x+4+2)=120
(x²+5x+4)²+2(x²+5x+4)=120
(x²+5x+4)²+2(x²+5x+4)-120=0
(x²+5x+4-10)(x²+5x+4+12)=0
x²+5x+4-10=0或x²+5x+4+12=0(无解)
x²+5x-6=0解得x1=-6,x2=1
即x1=-6,x2=1

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(x+1)(x+2)(x+3)(x+4)=120,
[(x+1)(x+4)]*[(x+2)(x+3)]=120,
(x^2+5x+4)(x^2+5x+6)=120,
(x^2+5x+4)[(x^2+5x+4)+2]=120,
(x^+5x+4)^2+2*(x^2+5x+4)-120=0,
[(x^2+5x+4)+12]*[(x^2+5x+4)-10]=0,<...

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(x+1)(x+2)(x+3)(x+4)=120,
[(x+1)(x+4)]*[(x+2)(x+3)]=120,
(x^2+5x+4)(x^2+5x+6)=120,
(x^2+5x+4)[(x^2+5x+4)+2]=120,
(x^+5x+4)^2+2*(x^2+5x+4)-120=0,
[(x^2+5x+4)+12]*[(x^2+5x+4)-10]=0,
(x^2+5x+16)(x^2+5x-6)=0,
因为x^+5x+16=(x+5/2)^2+39/4>0,
所以 x^2+5x-6=0,
(x+6)(x-1)=0,
x+6=0,或 x-1=0,
即 x=-6,或 x=1。

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4x+10=120
4x=110
x=27.5

(x+1)(x+2)(x+3)(x+4)=120
(x+1)(x+4)(x+2)(x+3)=120
(x^2+5x+4)(x^2+5x+6)=120
(x^2+5x)^2+10*(x^2+5x)+24-120=
(x^2+5x)^2+10(x^2+5x)-96=0
(x^2+5x+16)(x^2+5x-6)=0
(x^2+5x+16)(x+6)(x-1)=0
x=-6或 x=1

[(x+1)(x+4)][(x+2)(x+3)]=120~~~~(x*x+5x+4)(x*x+5x+6)=120~~~(x*x+5x+4)^2+2(x*x+5x+4)-12*10=0~~~~(x*x+5x-6)(x*x+5x+16)=0 ~~~~~~x=1或 x=-6

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