12x^2-10xy 2y^2 11x-5y 2

12分之x+10=7X分之x+2解12分之x+10=7分之1X+212分之x=7分之1X+2-1012分之x=7分之1X-8

x/(2x^2-11x-21)=(x+7)/(x^2-12x+35)x/(2x+3)(x-7)=(x+7)/(x-5)(x-7)x/(2x+3)=(x+7)/(x-5)x(x-5)=(x+7)(2x+

=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

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5x/(x²+x-6)+(2x-5)/(x²-x-12)=(7x-10)/(x²-6x+8)5x/(x-2)(x+3)+(2x-5)/(x+3)(x-4)=(7x-10)

(5-2x)/(x^2-x-12)=5x/(x^2+x-6)+(10-7x)/(x^2-6x+8)(5-2x)/(x-4)(x+3)=5x/(x+3)(x-2)+(10-7x)/(x-2)(x-4)分

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)

主要是考察因式分解这方面的技能啊![(2x²-5x+2)/(6x²+13x+6)]*[(6x²-5x-6)/(6x²-17x+10)]*[(12x²+

x/（2x^2-11x-21）=（x+7）/(x^2-12x+35)x/（2x+3）(x-7)=（x+7）/(x-5)(x-7)约去x-7(显然x=7不是方程的根)x/（2x+3)=（x+7）/(x-

5x/(x²+x-6)+(2x-5)/(x²-x-12)=(7x-10)/(x²-6x+8)5x/[(x+3)(x-2)]+(2x-5)/[(x-4)(x+3)]=(7x

就这么算啊(x+2)(x-12)+(x+7)(x+8)+(x+5)(x-10)=(x²-10x-24）+（x²+15x+56）+（x²-5x-50)=3x²-1

(X+2)(X+5)(X-4)(X+2)(y-3)(y-4)(x+9)(x-2)

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)

(12x-10)/21+(7x-9)/20=(2-x)/15+(8x-9)/1220(12x-10)+21(7x-9)=28(2-x)+35(8x-9)240x-200+147x-189=56-28x

=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

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

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

3x²=10x+83x²-10x-8=0x=(10±√(100+4×3×8)/2*3x=(10±√196)/6x=(10±14)/6∴x1=4x2=-2/33x²-10x