(10x+x+2)^2-[x+10(x+2)]=138怎样解

(x^2-x)^2-18(x^2-x)+72(x+1)(x+2)-x-10 (x+12)(x-2)+(x+7)(x+8)-(x+5)(x-10),其中x=2 (x-1)/(x+3x+2)+6/(2+x-x)-(10-x)/(4-x) 因式分解(x+1)(x+2)-x-10(x²-x)²-18(x²-x)+72 解方程(x/x(x+2))+(x/(x+2)(x+4))+.+x/(x+8)(x+10)=1 X+X+X+(X-1)+(X-2)+(X-3)+(X-4)=10(X-3)+(X-4) x^2+10x( )=(x+ )^2 x^2-10x=2-x (X-2)(X+3)+3x+10 2x×x=10,求x x+2x×10=10x+2x+36 x^2+2x/3x^2+x-10化简 x-10+X+20+X/2+2X=370 因式分解（x^2-x)-7x^2+7x+10 x+10+x-20+2x+x/2=370 因式分解(x^2-x)(x^2-x-10)-24 x/x^2-25-5/x^2+10x+25 (x^+x)^2-10(x^2+x)+25