Question

（x^2-9）(x+3分之1-x-3分之1-1)=?a- b分之b^2-(2a^2b-2ab^2)除以(a^2-2ab+

Answer 1

（x^2-9）(x+3分之1-x-3分之1-1)=(x^2-9)[((x-3-x-3)-(x^2-9))/x^2-9]=3-x^2

b^2/(a-b)-(2a^2b-2ab^2)/(a^2-2ab+b^2)+(a+b)a^2/(a^2-b^2)=b^2/(a-b)-[2ab(a-b)]/(a-b)^2+(a+b)a^2/(a^2-b^2)=b^2/(a-b)-2ab/(a-b)+a^2/(a-b)=(b^2-2ab+a^2)/(a-b)=(a-b)^2/(a-b)=a-