9x 6x平方-3(x-3分之2x平方)x=-2
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3x^2+xy-2y^2=(3x-2y)(x+y)=0x=-2y/3,或x=-y但由于你要求的式子中分母里有X平方-Y平方x=-y舍去其他就代入吧
你可以发个图给个题目吗,这样不知那些是分子,那些是第二项a再问:好吧再问:再答:再答:图中的第二题,你们是一个班的吧?问的问题都一样再问:应该不是再问:第4题就两个步骤?再答:是的再问:最后一步是x-
(3-x分之x-1)平方÷(x平方-6x+9分之9-x平方)平方*x平方-2x+1分之6+2x=[(x-1)/(3-x)]²÷[(9-x²)/(x²-6x+9)]
=(x+2)(x-1)/(x-3)×(x-3)²/(x+2)=(x-1)(x-3)=x²-4x+3
2x^2-6x+3=02x-6+3/x=02x+3/x=6平方(2x+3/x)^2=364x^2+9/x^2+12=364x^2+9/x^2=24
x的平方-9分之x的平方+2x-3乘x的平方-1分之x的平方-5x+6=(x-3)分之(x-1)乘[(x-1)(x+1)]分之(x-2)(x-3)=(x+1)分之(x-2)
)【(x²-2x+1)/(x²-1)】÷【(x-1)/(x²+x)】=【(x-1)²/(x+1)(x-1)】÷【(x-1)/x(x+1)】=【(x-1)/(x+
x-3分之5-x方-9分之x-2+2x-6分之x-1=5/(x-3)-(x-2)/(x^2-9)+(x-1)/2(x-3)=(10(x+3)-2(x-2)+(x-1)(x+3))/2(x+3)(x-3
原式=3(X-3)/[2(X-3)(X+2)]-(X-3)/[(X-2)(X-3)]+2X/[(X+3)(X-3)]=3/[2(X+2)]-1/(X-2)+2x/[(X+3)(X-3)]=(X-10)
1:2分之3X的平方-{-2分之1X平方}+{-2X平方}=3x²/2+x²/2-2x²=4x²/2-2x²=2x²-2x²=02
x平方-6x+8=x²-2×3x+9-9+8=(x-3)²-1=(x-3-1)(x-3+1)=(x-4)(x-2)x平方+x-6=(x-2)(x+3)12+x-x平方=(4-x)(
x的平方-2x+1分之x-3÷x-1分之x的平方-9=[(x-3)/(x-1)²]*[(x-1)/(x+3)*(x-3)]=1/(x-1)(x+3)
2x/3=x²/12+3/x²+4/xx²/12+3/x²=2x/3-4/xx²/4+9/x²=2x-12/x(x/2-3/x)²
(x²+6x+9)分之(x-3)+(9-x²)分之x=(x²+6x+9)分之(x-3)-(x²-9)分之x=(x+3)²分之(x-3)-[(x+3)(
2/(x+3)(x-3)+(x-2)/x(x-3)=1/x(x+3)两边乘x(x+3)(x-3)2x+x^2+x-6=x-3x^2+2x-3=0(x+3)(x-1)=0x+3在分母,不等于0所以x=1
x平方-x平方=0
原式=1/(x+1)*[(x-3)/(x+3)^2+2/(x+3)]-x/(x+3)(x-3)=1/(x+1)[3(x+1)/(x+3)^2]-x/(x+3)(x-3)=3/(x+3)^2-x/(x+
=(x+2)/(x-3)×(x-3)²/(x+2)(x-2)=(x-3)/(x-2)
原式=1/(x-3)-1/2(x+3)-x/(x-3)²=[2(x+3)(x-3)-(x-3)²-2x(x+3)]/[2(x+3)(x-3)²]=(2x²-18
解(3x²-9x-12)/(x²+2x+1)=3(x²-3x-4)/(x+1)²=3(x-4)(x+1)/(x+1)²=3(x-4)/(x+1)