解方程:x-1⼀x+1+x-4⼀x+4=x-2⼀x+2+x-3⼀x+3

2024-11-29 04:49:11
推荐回答(3个)
回答1:

解:
x-1/x+1+x-4/x+4=x-2/x+2+x-3/x+3
[(x-1)(x+4)+(x-4)(x+1)]/(x+1)(x+4)=[(x-2)(x+3)+(x-3)(x+2)]/(x+2)(x+3)
(2x²-8)/(x+1)(x+4)=(2x²-12)/(x+2)(x+3)
(x²-4)/(x²+5x+4)=(x²-6)/(x²+5x+6)
(x²+5x+4)(x²-6)=(x²-4)(x²+5x+6)
x⁴-6x²+5x³-30x+4x²-24=x³+5x³+6x²-4x²-20x-24
-6x²-30x+4x²=6x²-4x²-20x
-2x²-30x=2x²-20x
4x²+10x=0
2x(2x+5)=0
x=0或x=-5/2
经检验:x=0或x=-5/2是原方程的根

回答2:

(x-1)(x+2)(x+3)(x+4) + (x-4)(x+1)(x+2)(x+3) = (x-2)(x+1)(x+3)(x+4) + (x-3)(x+1)(x+2)(x+4)
(x+2)(x+3)[(x-1)(x+4)+(x+1)(x-4)] =(x+1)(x+4)[(x-2)(x+3)+(x+2)(x-3)]
(x²+5x+6)(x²+3x-4+x²-3x-4)=﹙x²+5x+4﹚﹙x²+x-6+x²-x-6﹚
(x²+5x+6)(2x²-8)=(x²+5x+4)(2x²-12)
(x²+5x+4)﹙2x²-8﹚+2(2x²-8)-(x²+5x+4)(2x²-12)=0
4(x²+5x+4) +4(x²-4)=0
4x²+20x+16+4x²-16=0
8x²+24x =0
x²+3x =0
x=0 或 x=-3

回答3:

(1)原题为:、4/x^2+1/x=2/(x-2)
通分:4(x-2)/x^(x-2)+x(x-2)/x^2(x-2)=2x^2/x^2(x-2),
取分子为0,则:
4(x-2)+x*(x-2)=2x^2.
x^2-2x+8=0.
判别式△=(-2)^2-4*1*8=-28,判别式△<0,∴所给方程在实数范围内无解;
(2)
若原题为:4/x^2+1/x=(2/x)-2.
通分:4/x^2+x/x^2=(2x/x^2)-2x^2/x^2.
取分子=0,则:
4+x=2x-2x^2.
2x^2-x+4=0.
判别式△=(-1)^2-4*2*4=-31<0,
同样原题在实数范围内无解。