Question

（x-1分之x^2+3x+2）+（6分之2+x-x^2）-(10-x分之4-x^2）,十万火急,明天就要交,

Answer 1

（x-1分之x^2+3x+2）+（6分之2+x-x^2）-(1-x分之4-x^2）,

=(x²+3x+2)/(x-1)+(2+x-x²)/6-(x²-4)/(x-1)

=(x²+3x+2-x²+4)/(x-1)+(2+x-x²)/6

=(3x+6)/(x-1)+(2+x-x²)/6

=(18x+36+2x-2+x²-x-x³+x²)/(6x-6)

=(34+19x+2x²-x³)/(6x-6)

题目肯定是错的