计算(x的平方-2x分之X 2-x的2次方-4x 4)

x^2/(x-3)-(2x-3)/(3-x)+(2x+6)/(3-x)=x^2/(x-3)+(2x-3)/(x-3)-(2x+6)/(x-3)=[x^2+(2x-3)-(2x+6)]/(x-3)=(x

[(3-x)/(x-1)-1]/[(x-2)/(x^2-2x+1)]=[(3-x-x+1)/(x-1)]*[(x^2-2x+1)/(x-2)}=[(-2x+4)/(x-1)]*[(x^2-2x+1)/

=1/(x-5)-6/(x+5)(x-5)-(x+1)/2(x+5)=2(x+5)/2(x+5)(x-5)-12/2(x+5)(x-5)-(x+1)(x-5)/2(x+5)(x-5)=(2x+10-1

1、(x+1)/(x-2)2、?3、同学,括号!

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-

x^2-3x+2=0(x-2)(x-1)=0x=2或x=1当x=2时x^2+1/x^2=2^2+1/2^2=4+1/4=17/4当x=1时x^2+1/x^2=1^2+1/1^2=1+1=2

x的平方-9分之1-x-3分之1除以x-1分之6+2x=1/(x^2-9)-1/(x-3)*(x-1)/2(3+x)=(2-(x-1))/2(x^2-9)=(3-x)/2(x^2-9)=-(x-3)/

x的平方-4分之x+1*（x+2）-x的平方+8x+16分之x的平方-16除以（x-4）=x的平方-四分之x+x+2-x的平方+8x+十六分之x的平方-(x-4分之16）=（x的平方-x的平方+十六分

你要求的是不是(（2x-6）/(x^2-4x+4))*(2x-4)/(x-3),如果是,(（2x-6）/(x^2-4x+4))*(2x-4)/(x-3)=（2(x-3)/(x-2)^2）*2(x-2)

原式=-12x2+5x+8用解二次方程的式子代入,得X1=0.5X2=-1/12

)【(x²-2x+1)/(x²-1)】÷【(x-1)/(x²+x)】=【(x-1）²/(x+1)（x-1)】÷【(x-1)/x(x+1)】=【(x-1）/(x+

298.3

=1/(x-5)-6/(x+5)(x-5)-(x+1)/2(x+5)=2(x+5)/2(x+5)(x-5)-12/2(x+5)(x-5)-(x+1)(x-5)/2(x+5)(x-5)=(2x+10-1

=[(x+2)/x(x-2)-(x-1)/(x-2)²]×(x-4)/x=[(x-2)(x+2)-x(x-1)]/x(x-2)²×(x-4)/x=(x-4)/x(x-2)²

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)

[(x-2)分之3x-2分之x]×x分之(x²-4)=[2(x-2)分之6x-2(x-2)分之(x²-2x)]×x分之[(x+2)(x-2)]=2(x-2)分之(4x-x²

题目等于{2X-(X+1)}/{(X-1)(X+1)}化简等于(x-1)分之一楼下的平方差公式怎么到最后的倒数第二步中（x+1）不见了?

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

x-y分之2x的平方-y-x分之x的平方-4xy+x-y分之2y的平方-x的平方=2x/(x-y)-(x²-4xy)/(y-x)+(2y-x²)/(x-y)=(2x+x²