2x+6=(x+3)²这道题怎么用分解因式法解

xy²/(x²y-y) × x²/(x²+x)=(x²-3x)/(x²-5x) × 2x-10/(x²-6x+9)=化简求植x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1 两个初升高衔接教材的数学题.①(x²+x+1)(x²+x+2)-12=(x²+x+1)(x²+x+1+1)-12=(x²+x+1)²+(x²+x+1)-12=(x²+x-3)(x²+x+5)[十字相乘法]②(x²+x)²-8(x²+x)+12=(x²+x-2)(x²+x-6)[ 已知2x²+2x-3=0 求代数式x(x²-x-1)+2(x²+1)-1/3x(3x²+6x) 已知2x²+3y²=6x 则x²+y²+2x的最大值 3x² - 9x + 2 = 0 x² - 14x = 8 x² + 6x = 1 x²-9/x²+2x-3乘3x²-x-2/x²-5x+6, x/y=2,求分式x²-xy+3y²/x²+xy+6y² (x² +x+1)/(x² +1)+(2x² +x+2)/(x² +x+1) =19/6化成(x² +x+1)/(x² +1)+(x² +1)/(x² +x+1)=13/6急嗷 化简x(x-1)+2x(x+1)-3x(2x-5)我是这一算的，化简x(x-1)+2x(x+1)-3x(2x-5)=x²-x+2x²+2x-6x²+15x=x+2x²-6x²+15x=18x+2x²-6x²=12x²-6x²可以相减吗，还是我算的不对了？ xy²/(x²y) - y×x²/(x²+x)=(x²-3x)/(x²-5x×2x) - 10/(x²-6x+9)=2a/(a²-4) - 1/(a-2)=2/(1-a ) - 1/(a-1)=x²-1/(x²-x-2)除以x/2x-4,其中x=1/2[x-x/(x+1)]除以[x/(2x-4)] ,其中x=√2+1 6x²-xy-15y²=0 求x²-2xy+4y²/2x²-3xy-3y²告诉下 6x²-xy-15y²=0 求2x²-3xy-3y²分之x²-2xy+4y² 若x²+xy-2y²=0,则x²+3xy+y²/x²+y² x²+xy-2y²=0则x²+3xy+y²/x²+y² 1.x²-2x-3 2.x²-5x-6 3.x²-9x-10 4.x²-9x-10 5.x²-2x-8 6.x²+2x-8 7.x²-x-12 8.x²+x-12 9.x²+6x-16 10.x²-14x-32 11.x²+7x-18 [()+6x-7]-[5x²+()-()]=3x²-2x+5 (-x+2x²+5)+(-3+4x²-6x)=? 问几道有关分式的数学题 ,(1)[(5xy)/(x²+xy)]*[(xy+y²)/(2x²y²)]^3(2)[(x^3-3x+4)/(x²+2x-3)]÷[(x²-16)/(x²+6x+9)](3)(x²+x-3)/(x²+x-2)+1=(2x²+4x+1)/(x²+2x+1)(4)已知a²+4a+1=0 ,且(a^4+