求证(x^2-9)(x-7)(x-1)
|X-1|+|X-2|+|X-3|+|X-4|+|X-5|+|X-6|+|X-7|+|X-8|+|X-9|+|X-10|
求证:2(根号x)>3-(1/x) (x>1)
y=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10)的导数在x=1
设函数f(x)=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)(x-10),
x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)谁会解这道方程题
分式方程.x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)
x/(x-2)+(x+9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)怎样解这道方程
解方程:x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)
x+2/x+3-x+1/x+2=x+8/x+9-x+7/x+8
方程x+1x+2+x+8x+9=x+2x+3+x+7x+8
(1) x-3/x-2 - x-5/x-4=x-7/x-6 - x-9/x-8
(4x+2)/x+(4x-22)/(x-5)=(x-6)/(x-4)+(7x+9)/(x+1)