解方程:(x-1)(x-2)(x+2)(x+3)=60

来源:学生作业帮助网 编辑:作业帮 时间:2024/11/15 06:19:24

解方程:(x-1)(x-2)(x+2)(x+3)=60
解方程:(x-1)(x-2)(x+2)(x+3)=60

解方程:(x-1)(x-2)(x+2)(x+3)=60
数学之美团为你解答
(x-1)(x-2)(x+2)(x+3)=60
[(x-1)(x+2)] [(x-2)(x+3)] = 60
(x²+x-2)(x²+x-6)=60
(x²+x)²-8(x²+x)+12=60
(x²+x)²-8(x²+x)-48=0
(x²+x-12)(x²+x+4)=0
x²+x-12= 0 或 x²+x+4=0
x²+x-12= 0 即 (x+4)(x-3)=0,得x₁=-4;x₂=3;
x²+x+4=0 ,Δ

解方程:
(x-1)(x-2)(x+2)(x+3)=60
(x^2-4)(x-1)(x+3)=5*2*6
那么:x^2-4=5
x1=3 x2=-3 (不合题意,舍去)
x-1=2
x3=3
x+3=6
x4=3
即:x=3

解方程:(x-1)(x-2)(x+2)(x+3)=60
[(x-1)(x+2)][(x-2)(x+3)]=(x²+x-2)(x²+x-6)=(x²+x)²-8(x²+x)+12=60
即有(x²+x)²-8(x²+x)-48=(x²+x-12)(x²+x+4)=0
由x...

全部展开

解方程:(x-1)(x-2)(x+2)(x+3)=60
[(x-1)(x+2)][(x-2)(x+3)]=(x²+x-2)(x²+x-6)=(x²+x)²-8(x²+x)+12=60
即有(x²+x)²-8(x²+x)-48=(x²+x-12)(x²+x+4)=0
由x²+x-12=(x+4)(x-3)=0,得x₁=-4;x₂=3;
由x²+x+4=0,得x₃=(-1+i√15)/2;x₄=(-1+i√15)/2;

收起