0.4x-12=4/1x这道题怎么解

是这样的：x^5+x^4=x^3(x^2+x)=(x^2+x)[(x^3-1)+1]=(x^2+x)(x^3-1)+x^2+x=[x(x+1)(x-1)](x^2+x+1)+x^2+x=(x^3-x)

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

题目有点问题,应是(x²+3x+1)/(4x²+6x-1)-(12x²+18x-3)/(x²+3x+1)=2(x²+3x+1)/(4x²+6

(x+2-12/x-2)/(4-x)/(x-2),其中x=-1/2=(x²-4-12)(x-2)÷(4-x)/(x-2)=(x²-16)/(4-x)=-(4-x)(4+x)/(4-

两边同时乘以x(x+4)(x+12)4x(x+12)=3(x+4)(x+12)+x(x+4)4x²+48x=3(x²+16x+48)+x²+4x144+4x=0得x=-3

(x/x(x+2))+(x/(x+2)(x+4))+.+x/(x+8)(x+10)=(x/2)*(2/x(x+2))+(2/(x+2)(x+4))+.+2/(x+8)(x+10)=(x/2)*[1/x

第一道题设t=x²-x；第二道题设t=x²/x+1；第三道题你没括号我不知道什么意思应该和第一道题一样；第四个的第一个东西不是可以约掉么?..是不是弄错了,如果没弄错那就是设t=x

即x²-5x=-1所以原式=x³-5x²+x²-5x+x-1=x(x²-5x)+(x²-5x)+x-1=-x-1+x-1=-2

解(1)x-0.75x=120.25x=121/4x=12x=12×4x=48(2)2.8x+x=7.63.8x=7.6x=7.6÷3.8x=2(3)2.36x+2.64x=185x=18x=18÷5

每个绝对值里都是正数所以原式=x+1+x+3+x+5+……+x+13-x-2-x-4-x-6-……-x-12=x-1*6+13=x+7=π/2+7

=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

题目应是1/(x^2+x﹚+1/(x^2+3x+2﹚+1/(x^2+5x+6﹚+1/(x^2+7x+12﹚=4/211/x(x+1﹚+1/(x+1)(x+2)+1/(x+2)(x+3)+1/(x+3)

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

1/6x+1/12x+1/7x+5+1/2x+4=x2/12x+1/12x+6/12x-12/12x+1/7x=-9-1/4x+1/7x=-9-7x+4x=-9*28-3x=-9*28x=84

1/2x^2-1/4x+0.2x^3+0.25x-0.5x^2-1/5x^3+0.5x-1=(1/2x^2-0.5x^2)+(0.25x-1/4x)+(0.2x^3-1/5x^3)+0.5x-1=0.

x-(1/5+1/5)x-39=39x-2/5x=39+393/5x=78x=78÷3/5x=130x-1/4x-2/5x=112(1-1/4-2/5)x=1127/20x=112x=112÷7/20

X-1=3X+32X=-4;X=-2x-3=12-4x5X=15；x=36x-14=5x-10x+1=2-4x5x=1;x=0.2

=x^2002(x^4+x^3+x^2+x+1)=0提取公因式就行了

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