先化简[2x平方+2x分之x平方-1]-搜答案-搜搜做题作业网

解x/(x+2)-(x²+2x+1)/(x+2)÷(x²-1)/(x-1)=x/(x+2)-(x+1)²/(x+2)÷(x-1)(x+1)/(x-1)=x/(x+2)-(

x+2分之x平方-2x-x-1分之x平方-4X+4除以x平方-16分之x平方+4x=x(x-2)/(x+2)÷(x-2)^2/(x-1)÷[x(x+4)/(x+4)(x-4)]=[x(x-2)/(x+

x+2分之x-x+2分之x平方+2x+1除以x-1分之x平方-1=[x/(x+2)-(x²+2x+1)/(x+2)]÷(x²-1)/(x-1)=[(x-x²-2x-1)/

原式=x³+3x+x³-3x²-3x³+3x²+3x=-x³+6x=-1/8+3=23/8

（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)]

=2x/x²-(2x-1)/x²=(2x-2x+1)/x²=1/x²

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

原式=[(x²+x-12)/(2-x-x²)]²×[(x²-3x+2)/(x²+5x+4)]²÷[(x²-5x+6)/(x

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）（

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

0.5（x+1）的平方

[（x-1)/x-(x-2)/(x+1)]÷[(2x^2-x)/(x^2+2x+1)]=[(x-1)(x+1)-x(x-2)]/x(x+1)÷[x(2x-1)/(x+1)^2=(2x-1)/x(x+1

=[(x+2)/x(x-2)-(x-1)/(x-2)²]×x/(4-x)=[(x²-4-x²+4)/x(x-2)²]×x/(4-x)=1/(x-2)²

=x/(x+2)(x-2)-(x-2)/(x+2)(x-2)=[x-(x-2)]/(x+2)(x-2)=2/(x²-4)

(x-5分之x-5-x分之x)\x的平方-25分之2x=2x/(x-5)*(x-5)(x+5)/2x=x+5

①X平方+2X分之X+2-X平方-4X+4分之X-1=(x+2)/x(x+2)-(x-1)/(x-2)²=[(x-2)²-x²+x]/x(x-2)²;=(4-3

代数式的值应该与x的取值无关,即,代数式化简后为一个常数证明：(x的平方-1)(x的平方-x)分之(x的平方+x)(x的平方-2x+1)＝[x(x+1)(x-1)(x-1)]/[(x+1)(x-1)x

4-4x分之9x-9=9(x-1)/4(1-x)=-9/4

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

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