作业帮(10x+x+2)^2-[x+10(x+2)]=138怎样解
来源：学生作业帮助网 编辑：作业帮 时间：2024/05/13 12:16:47
![(10x+x+2)^2-[x+10(x+2)]=138怎样解](/uploads/image/z/327292-52-2.jpg?t=%2810x%2Bx%2B2%29%5E2-%5Bx%2B10%28x%2B2%29%5D%3D138%E6%80%8E%E6%A0%B7%E8%A7%A3)
(10x+x+2)^2-[x+10(x+2)]=138怎样解
(10x+x+2)^2-[x+10(x+2)]=138怎样解
(10x+x+2)^2-[x+10(x+2)]=138怎样解
10x+(x+2)]-[10(x+2)+x]=138 (11x+2)-(11x+20)=138 121x+33x-154=0 11x+3x-14=0 (11x+14)(x-1)=0 x>0 x=1 所以原来的数是31
(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