Đáp án:
$a)x=\dfrac{7}{4}$
Giải thích các bước giải:
$a)\left ( \dfrac{28}{2017}+\dfrac{20}{2019}-\dfrac{16}{2021} \right )x=\dfrac{98}{4034}+\dfrac{70}{4038}-\dfrac{56}{4042}\\
\Leftrightarrow \left ( \dfrac{28}{2017}+\dfrac{20}{2019}-\dfrac{16}{2021} \right )x=\dfrac{49}{2017}+\dfrac{35}{2019}-\dfrac{28}{2021}\\
\Leftrightarrow 4.\left ( \dfrac{7}{2017}+\dfrac{5}{2019}-\dfrac{4}{2021} \right )x=7.\left (\dfrac{7}{2017}+\dfrac{5}{2019}-\dfrac{4}{2021} \right )\\
\Leftrightarrow x=\dfrac{7.\left (\dfrac{7}{2017}+\dfrac{5}{2019}-\dfrac{4}{2021} \right )}{4.\left ( \dfrac{7}{2017}+\dfrac{5}{2019}-\dfrac{4}{2021} \right )}=\dfrac{7}{4}$
$b)\dfrac{3n+2}{2n+1}$
Gọi $UCLN(3n+2;2n+1)$ là d
$\Rightarrow {\left\{\begin{aligned}3n+2\vdots d\\ 2n+1\vdots d\end{aligned}\right.}\\
\Rightarrow {\left\{\begin{aligned}6n+4\vdots d\\ 6n+3\vdots d\end{aligned}\right.}\\
\Rightarrow (6n+4)-(6n+3)\vdots d\\
\Rightarrow 6n+4-6n-3\vdots d\\
\Rightarrow 1\vdots d\\
\Rightarrow d\in U(1)={1}$
Vì $d=1$ hay $UCLN(3n+2;2n+1)=1$
$\Rightarrow \dfrac{3n+2}{2n+1}$ là phân số tối giản
