Bài 1.
$h=\frac{2}{3}\times 6=4cm$
Kẻ AH, CK vuông góc với BD
${{S}_{ABD}}=\frac{AB\times h}{2}=\frac{6\times 4}{2}=12c{{m}^{2}}$
${{S}_{CBD}}=\frac{CD\times h}{2}=\frac{10\times 4}{2}=20c{{m}^{2}}$
Tam giác ABD và CBD có chung đáy BD, nên: $\frac{{{S}_{ABD}}}{{{S}_{CBD}}}=\frac{AH}{CK}=\frac{12}{20}=\frac{3}{5}$
Tam giác AOB và BOC có chung đáy OB, nên: $\frac{{{S}_{AOB}}}{{{S}_{BOC}}}=\frac{AH}{CK}=\frac{3}{5}$
Mà AOB và BOC lại có chung chiều cao hạ từ B nên: $\frac{{{S}_{AOB}}}{{{S}_{BOC}}}=\frac{OA}{OC}=\frac{3}{5}$
Bài 2.
$\frac{2019}{2020}+\frac{2020}{2019}=1-\frac{1}{2020}+1+\frac{1}{2019}=2+\left( \frac{1}{2019}-\frac{1}{2020} \right)$
Vì: $\frac{1}{2019}>\frac{1}{2020}$ nên $\frac{1}{2019}-\frac{1}{2020}>0\Rightarrow 2+\left( \frac{1}{2019}-\frac{1}{2020} \right)>2$
$\frac{444443}{222222}=2-\frac{1}{222222}\frac{444443}{222222}$
Bài 3.

$\begin{align}
& A=\frac{4}{7\times 31}+\frac{6}{7\times 41}+\frac{9}{10\times 41}+\frac{7}{10\times 57} \\
& A\times \frac{1}{5}=\frac{4}{31\times 35}+\frac{6}{35\times 41}+\frac{9}{41\times 50}+\frac{7}{50\times 57} \\
& A\times \frac{1}{5}=\frac{35-31}{31\times 35}+\frac{41-35}{35\times 41}+\frac{50-41}{41\times 50}+\frac{57-50}{50\times 57} \\
& A\times \frac{1}{5}=\frac{1}{31}-\frac{1}{35}+\frac{1}{35}-\frac{1}{41}+\frac{1}{41}-\frac{1}{50}+\frac{1}{50}-\frac{1}{57} \\
& A\times \frac{1}{5}=\frac{1}{31}-\frac{1}{57} \\
& A=\left( \frac{1}{31}-\frac{1}{57} \right)\times 5=\frac{5}{31}-\frac{5}{57} \\
\end{align}$
$\begin{align}
& \frac{5}{57}\times x=\frac{5}{31}-\frac{5}{57} \\
& x=\left( \frac{1}{31}-\frac{1}{57} \right):\frac{1}{57} \\
& x=\frac{26}{31} \\
\end{align}$