（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,

（x-1/x^2+3x+2）+6/(2+x-x^2）-（10-x/4-x^2）
=(x-1)/(x²+3x+2)+6/(2+x-x²)-(10-x)/(4-x²)
=(x-1)/(x+1)(x+2)-6/(x-2)(x+1)-(x-10)/(x-2)(x+2)
=[(x-1)(x-2)-6(x+2)-(x-10)(x+1)]/(x+1)(x+2)(x-2)
=(x²-3x+2-6x-12-x²+9x+10)/(x+1)(x+2)(x-2)
=0/(x+1)(x+2)(x-2)
=0