(1-x)9=a0 a1X a2X

等于37因为：【】都是一样的数字的话那他们就可以合并同类项：[]{10+1}*5=1980所以算出为37~

1/[x(x-1)]=1/(X-1)-1/X1/[x(x+1)]=1/X-1/(X+1)1/(X-1)-1/X+1/X-1/(X+1)+...+1/(X+9)-1/(X+10)=1/(x+10)1/(

原式=x³-6x²-9x-2x²+12x+18-（x²-5x）（x-3）=x³-8x²+3x+18-（x³-3x²-5x

设a=(x-1)(x-2)(x-3)(x-4)(x-5)(x-6)(x-7)(x-8)(x-9)那么y=a*(x-10);那么y^=a^*(x-10)+a*(x-10)^=a^*(x-10)+a那么y

1）（x-3）/（x-2）-（x-5）/（x-4）=（x-7）/（x-6）-（x-9）/（x-8）化简得【（x-3）（x-4）-（x-5）/（x-2）】/【（x-2）（x-4)】=【（x-7）（x-8

 再问：额、不懂再答： 再答：后面的看做一个整体再问：好的吧、谢谢大神再答：回来的话，请采纳再问：啊、突然明白了呢。。。

1x+2x+3x+4x+5x+6x+7x+8x+9x+10x+11x+12x+13x+14x+15x=550120x=550x=55/12=4.583

X(1+2+3+4+5+6+7+8+9+10+11)=1980,X66=1980等于30

x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)1+2/(x-2)+1-2/(x-7)=1+2/(x-1)+1-2/(x-6)1/(x-2)-1/(x-7)=1/

1/x(x+1)+1/(x+1)(x+2)+...+1/(x+9)(x+10)=1/(x+10),1/x(x+1)+1/(x+1)(x+2)+...+1/(x+9)(x+10)=1/x-1/(x+1)

(x-9)/(x-7)-(x+1)/(x-1)=(x-8)/(x-6)-x/(x-2)[(x-9)(x-1)-(x+1)(x-7)]/[(x-7)(x-1)]=[(x-8)(x-2)-x(x-6)]/

/>(4X+2)/X+【4(X-5)-2】/(X-5)=【(X-4)-2】/(X-4)+【7(X+1)+2】/(X+1)4+2/X+4-2/(X-5)=1-2/(X-4)+7+2/(X+1)8+2/X

x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)1+2/(x-2)+1-2/(x-7)=1+2/(x-1)+1-2/(x-6)1/(x-2)-1/(x-7)=1/

(x+2)/(x+3)-(x+1)/(x+2)=(x+8)/(x+9)-(x+7)/(x+8)(x+2)²/[(x+2)(x+3)]-(x+1)(x+3)/[(x+2)(x+3)]=(x+8

同分化简得32x+128/(x+1)(x+3)(x+5)(x+7)=0所以得x=-4给分吧算了好久的····

应该是-2x²,不是-3x²3x²-x=1原式=9x^4-3x³+15x³-5x²+3x²-7x+2001=3x²(3x

1.9/x+x=6,x^2+9=6x,x^2-6x+9=0,(x-3)^2=0,x=3.2.(x+1)/x+2x/(x+1)=3两边同乘以x（x+1)2x^2+(x+1)^2=3*x*(x+1)2x^

X^2+X-7=0X1=根29-1X2=-1-根29X^2+X-2=0X1=-2X2=19-6X+X^2+X=9X^2-5X=0X(X-5)=0X1=0X2=5

2x/x^2-9=x-1/x^2-6x+9-1/x+32x/(x+3)(x-3)=(x-1)/(x-3)²-1/(x+3)两边同乘以(x-3)²(x+3)2x(x-3)=(x-1)

x/(x-2)+(x-9)/(x-7)=(x+1)/(x-1)+(x-8)/(x-6)1+2/(x-2)+1-2/(x-7)=1+2/(x-1)+1-2/(x-6)1/(x-2)-1/(x-7)=1/