要使(x^2 mx 8)(9x^2-3x n)

|x-1|+|x-10|表示数轴上x到1的距离+x到10的距离.显然最小值是9,此时x只要在1到10之间就好.类似的,|x-2|+|x-9|的最小值是7,此时x在2到9之间就好.|x-3|+|x-8|

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。。。

（X²+3X+9/X²-27)+(6X/9X-X²)-(X-1/6+2X)=X²+3X+9/X²-27+6/9-X²-X+1/6+2X=9/

x(1+2+...+9)=x(9-8-7-...-1)x=0记得采纳啊

是9x-x³原式=(x²+3x+9)/(x-3)(x²+3x+9)-6x/x(x-3)(x+3)-(x-1)/2(x+3)=1/(x-3)-6/(x-3)(x+3)-(x

中间错了原式=(x+2)/(x-3)(x+2)+2x/x(x-3)-(x+1)/(x-3)²=1/(x-3)+2/(x-3)-(x+1)/(x-3)²=3/(x-3)-(x+1)/

原式=[x/x(x-3)](x+3)(x-3)=(x+3)(x-3)/(x-3)=x+3

(2x-6)/(x²-4x+4)÷(3-x)/(x²+6x+9)×(2-x)/(x+3)=2（x-3)/(x-2)^2X(x+3)^2/(3-x)X(2-x)/(x+3)=2（x-

原式=(x+2)/(x+2)(x-3)-2x/x(x-3)-(x+1)/(x-3)²=1/(x-3)-2/(x-3)-(x+1)/(x-3)²=-1/(x-3)-(x+1)/(x-

∫2^x*3^x/(9^x-4^x)dx=∫(2/3)^xdx/[1-(4/9)^x]=[ln(2/3)]^(-1)∫d[(2/3)^x]/{1-[(2/3)^x]^2}={[ln(2/3)]^(-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

原式=[x(x-3)/(x-3)²][(x-5)(x-6)/x(x-5)]-1=[x/(x-3)][(x-6)/x]-1=(x-6)/(x-3)-1=(x-6-x+3)/(x-3)=-3/(

(x+2)/(x^2-x-6)+2x/(x^2-3x)-x/(x^2-6x+9)=(x+2)/[(x-3)(x+2)]+2x/[x(x-3)]-x/[(x-3)^2]=1/(x-3)+2/(x-3)-

我回答过的,看看、

化简：1/x(x+1)+1/(x+1)(x+2)+1/(x+2)(x+3)+...+1/(x+8)(x+9)=1/x-1/(x+1)+1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+.

（X²+3X+9/X²-27)+(6X/9X-X²)-(X-1/6+2X)=X²+3X+9/X²-27+6/9-X²-X+1/6+2X=9/

原式=[(2X-3)/X-X/X]/[(X+3)(X-3)/X]=[(2X-3-X)/X]/[(X+3)(X-3)/X]=[(X-3)/X]×[X/(X+3)(X-3)]=X(X-3)/[X(X+3)

答：结论是无解的设1和4中间的正方形边长为x则左边中间的正方形边长为x+1左下角边长为x+1+x=2x+1所以：右下角正方形边长2x+1+x-4=3x-3所以：最大的正方形底部边长=2x+1+3x-3

(x^2-9)/(x+3)=(x+3)(x-3)(x+3)=(x+3)²(x-3)