dx x^2(9-x^2)^1 2

|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|

1.dy/dx=(dy/dt)/(dx/dt)=(3t^2-3)/(2-2t)=-3/2*(t+1)2.a(t)=s''(t)=((e^t-(-1)e^-t)/2)'=(e^t-e^-t)/2=s(t

=x^18+x^15+x^12+x^9+x^6+x^3+1+x^3-1+5-4分解因式把下面的18换成21就是了=(x^18+1)+(x^15+1)+(x^12+1)+(x^9+1)+(x^6+1)+

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

x(1+2+...+9)=x(9-8-7-...-1)x=0记得采纳啊

(1/x^2+3x+2)+(1/x^2+5x+6)+(1/x^2+7x+12)+(1/x^2+9x+20)=1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)

x+9x=6x*2+1212x-x-9x=-122x=-12x=-12÷2x=-6x+9-x/3=4两边乘33x+9-x=123x-x=12-92x=3x=3÷2x=3/2

化简得2/x^2+6x=0,即（1+3x^3)/x^2=0,所以x=-（1/3)^(1/3)

∫xtan²xdx设u=x,dv=tg^2xdx,则du=dx,v=tgx-x于是∫xtan²xdx=x(tgx-x)-∫(tgx-x)dx=x(tgx-x)+Ln|cosx|+x

原式=3(X-3)/[2(X-3)(X+2)]-(X-3)/[(X-2)(X-3)]+2X/[(X+3)(X-3)]=3/[2(X+2)]-1/(X-2)+2x/[(X+3)(X-3)]=(X-10)

=9x^4+4(x+1)-3x^2-7x+1=9x^4-3x^2-3x+5=9x^4-3x(x+1)+5=9x^4-3x×3x^3+5=5

提示,设t=（x^2+3x)/(x^2+X-4)算出t再求x

原式=1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)(x+4)+1/(x+4)(x+5)=1/(x+1)-1/(x+2)+1/(x+2)-1/(x+3)+1/(x+3)-1/(x+

1/(x^2+3x+2)+1/(x^2+5x+6)+1/(x^2+7x+12)+1/(x^2+9x+20)＝1/﹙x＋1﹚﹙x＋2﹚＋1/﹙x＋2﹚﹙x＋3﹚＋1/﹙x＋3﹚﹙x＋4﹚＋1/﹙x＋4﹚

①x8x1(x+8)(x+1)②x-3x-3(x-3)^2③x3x-4(x+3)(x-4)④x3x7(x+3)(x+7)⑤x5x-25(x+5)(x-25)⑥x12x3(x+1)(2x+3)

分步写,好让看的清楚符号[√(x^2-6x+9)/x^2-x-12]=√(x-3)^2/(x-4)(x+3）=（x-3)/(x-4)(x+3）；(x^3-16x)/(x^2-3x)=x(x^2-16)

9x-12x+5x=-3x+5x=2x3x-5x-2+5x+8=-2x+5x-2+8=3x+63x+5+2x+3-7-11x=3x+2x-11x+5+3-7=5x-11x+8-7=-6x+1

=x^2+10x-24+x^2+15x+56-x^2+5x+50=x^2+30x+82=4+60+82=146其实,不化简,直接代入=14*0+9*10+7*8=0+90+56=146也很简单

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

1/x^2-3x+2-1/x^2-5x+6-1/x^2-7x+12-1/x^2-9x+20=1/(x-1)(x-2)-1/(x-2)(x-3)-1/(x-3)(x-4)-1/(x-4)(x-5)=1/