Đáp án:
a) $\dfrac{4x-3}{6}+\dfrac{2x+3}{6}=x$
b) $\dfrac{x+2018}{x(x-3)}-\dfrac{2021}{x(x-3)}=\dfrac{1}{x}$
Giải thích các bước giải:
$a) \dfrac{4x-3}{6}+\dfrac{2x+3}{6}\\=\dfrac{4x-3+2x+3}{6}\\=\dfrac{6x}{6}=x\\b) \dfrac{x+2018}{x(x-3)}-\dfrac{2021}{x(x-3)}\\=\dfrac{x+2018-2021}{x(x-3)}\\=\dfrac{x-3}{x(x-3)}=\dfrac{1}{x}$