若8x 9 x² x-6=A x 3 B x-2

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

 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。。。

[1-1/(x-4)]+[1-1/(x-8)]=[1-1/(x-7)]+[1-1/(x-5)]1/(x-4)+1/(x-8)=1/(x-7)+1/(x-5)(x-4+x-8)/(x-4)(x-8)=(

x(1+2+...+9)=x(9-8-7-...-1)x=0记得采纳啊

x+x^2+x^3+x^4+x^5+x^6+x^7+x^8=（x+x^2+x^3+x^4）+（x^5+x^6+x^7+x^8）=x（1+x+x^2+x^3）+x^5（1+x+x^2+x^3）=（x+x

原式=x^4-8x^3+13x^2+10+2x(x^2-8x+13)+(x^2-8x+13)=x^2(x^2-8x+13)+10=10分组代入后得10

/>因为：x-5/x-6+x-8/x-9=x-7/x-8+x-6/x-7,所以：1+1/(x-6)+1+1/(x-9)=1+1/(x-8)+1+1/(x-7)即：1/(x-6）+1/(x-9)=1/(

即1/(x-5)+1/(x-9)=1/(x-6)+1/(x-8)(2x-14)[1/(x-5)(x-9)-1/(x-6)(x-8)]=01/(x-5)(x-9)-1/(x-6)(x-8)≠0所以2x-

1+x+x^2+x^3=0x+x^2+x^3+x^4+x^5+x^6+x^7+x^8=(x+x^2+x^3+x^4)+(x^5+x^6+x^7+x^8)=x(1+x+x^2+x^3)+x^5(1+x+

设x1=4-√3,x2=4+√3,是方程X^2-8X+13=0的两根所以X1^2-8X1+15=2X^4-6X^3-2X^2+18X+23=(X+1)^2*(X^2-8X+13)+10=10所以原式=

这道题不用那么复杂.先移项.(x-4)/(x-5)-(x-5)/(x-6)=(x-7)/(x-8)-(x-8)/( x-9)再通分.（x-4)(x-6)/(x-5)(x-6)-(x-5)(x

左右两边同乘以(X-5)(X-6)(X-8)(X-9)得到(X-4)(X-6)(X-8)(X-9)-(X-5)(X-5)(X-8)(X-9)=(X-5)(X-6)(X-7)(X-9)-(X-5)(X-

(x-4)/(x-5)-(x-5)/(x-6)=(x-7)/(x-8)-(x-8)/(x-9)1+1/(x-5)-1-1/(x-6)=1+1/(x-8)-1-1/(x-9)1/(x-5)-1/(x-6

7/x+8/x-4/x-5/x=-8-9+5+66/x=-6x=-1

你这题显然,打错了已知条件建议改为x=13/√(19+8√3)=13/√(4+√3)^2=13/(4+√3)=4-√34-x=√3两边平方x^2-8x+13=0(X^4-6X^3-2X^2+18X+2

x-4/x-5-x-5/x-6=x-7/x-8-x-8/x-91+1/x-5-1-1/x-6=1+1/x-8-1-1/x-91/x-5-1/x-6=1/x-8-1/x-9-1/(x-5)(x-6)=-

[(x^2-2x+4)/(√x^2-4x+4)]/[(x^3+8)/(x^2-4)]*[|6-x|/(x^2-5x-6)-[1/(x^2-x+1)]^-1=[(x^2-2x+4)/|x-2|]/[(x

(X-4)/(X-5)-(X-5)/(X-6)=(X-7)/(X-8)-(X-8)/(X-9)[(x-4)(x-6)-(x-5)2]/(x-5)(x-6)=[(x-7)(x-9)-(x-8)2]/(x

x=根号下19-8倍根号3=根号下[4^2+(根号3)^2-2*4*根号3]=根号(4-根号3)^2=4-根号3x-4=-根号3(x-4)^2=3x^2-8x+13=0x^2-8x+13=0,所以x^

(x-4)/(x-5)+(x-8)/(x-9)=(x-7)/(x-8)+(x-5)/(x-6)1+1/(x-5)+1+1/(x-9)=1+1/(x-8)+1+1/(x-6)1/(x-5)+1/(x-9