Question

分式的加减计算题 急(2a-b分之2a)+（b- 2a分之b）(x-2分之4)+（2-x分之x+2）（4a的平方分之b的

Answer 1

2a/(2a-b)+b/(b-2a)=2a/(2a-b)-b/(2a-b)=(2a-b)/(2a-b)=1

4/(x-2)+(x+2)/(2-x)=4/(x-2)-(x+2)/(x-2)=(2-x)/(x-2)=-1

b^2/4a^2-c/a=(b^2-4ac)/4a^2

1-1/(x+1)=x/(x+1)

1/r1+1/r2=(r1+r2)/r1r2

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

(a-1)/(a+1)-1/a=(a^2-a-a-1)/[a(a+1)]=(a^2-2a-1)/[a(a+1)]