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问题：几道分式方程的题解方程：1/(x-1)(X-2)=1/(X-4)(X-5)X/（X-1)-（X-1）/（X-2）=（x-3）/(x-4)-(x-4)/(X-5)
答案：↓↓↓ 宋培华的回答： 网友采纳 第一题: 1/(x-1)(x-2)=1/(x-4)(x-5) 1/x²-2x-x+2=1/x²-5x-4x+20 1/x²-3x+2=1/x²-9x+20 ∵分子相等,∴分母相等 x²-3x+2=x²-9x+20 6x=18 x=3 经检验,x=3是方程的根 第二题: x/（x-1)-（x-1）/（x-2）=（x-3）/(x-4)-(x-4)/(x-5) x/x-1-x-1/x-2=x-3/x-4-x-4/x-5 x(x-2)(x-4)(x-5)-(x-1)²(x-4)(x-5)/=(x-1)(x-2)(x-3)(x-5)/-(x-1)(x-2)(x-4)² (x²-2x)(x²-9x+20)-(x²-2x+1)(x²-9x+20)=(x²-3x+2)(x²-8x+15)-(x²-3x+2)(x²-8x+16) (x^4-11x³+38x²-40x)-(x^4-11x³+39x²-49x+20)=(x^4-11x³+41x²-61x+30)-(x^4-11x³+42x²-64x+32) x^4-11x³+38x²-40x-x^4+11x³-39x²+49x-20=x^4-11x³+41x²-61x+30-x^4+11x³-42x²+64x-32 -x²+9x-20=3x-x²-2 6x=18 x=3 经检验,x=3是方程的根 | 
