题目下列等式成立的是( )A.1a+2b=3a+b\dfrac{1}{a}+\dfrac{2}{b}=\dfrac{3}{a+b}a1+b2=a+b3B.abab−b2=aa−b\dfrac{ab}{ab-{{b}^{2}}}=\dfrac{a}{a-b}ab−b2ab=a−baC.12a+b=1a+b\dfrac{1}{2a+b}=\dfrac{1}{a+b}2a+b1=a+b1D.a−a+b=aa+b\dfrac{a}{-a+b}=\dfrac{a}{a+b}−a+ba=a+ba知识点：分式的基本性质,分式的加减章节：未标注
