1 2[2x-1 3(10-70)

原式=20x²-x-30-20x²+14x-26=13x-56=13×(-12/13)-56=-12-56=-68

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

1.x=22.x=43.x=64.第四题应该打错了吧,前面要是5x+5.那么x=7前面要是5x-5,那么x=5

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

⑴∵(1-2)x(2-3)x···x(9-10)=﹣1∴﹣(1-2)x(2-3)x···x(9-10)=1∴1x(10-11)x(11-12)x(12-13)x…x(99-100)=﹣(1-2)x(2

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

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

答案是10所有的多项式都可以消掉到最后只剩下了107x^3-10x^3=-3x^3-6x^3y+6x^3y=0直接消掉+3x^2y-3x^2y=0直接消掉3x^3和第二行剩的那个-3x^3也可以消掉整

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+

30-3（3-X）=4X30-9+3x=4x21=xx=213X+12*6=15X+1272-12=15x-3x12x=60x=54X-2(50-X)=204x-100+2x=206x=120x=20

1.7x=14x=22.4x-6-x+2=83x=12x=43.4x-2=22x=64.5x-30=5x=7

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)

原式=(x-1)(x^15+x^14+……+1)/(x-1)=(x^16-1)/(x-1)=(x^8+1)(x^4+1)(x²+1)(x+1)(x-1)/(x-1)=(x^8+1)(x^4+

=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