Gọi công thức của hợp chất là CxHyOz.

Khối lượng phân tử của hợp chất là 12.x + 1.y + 16.z = 60. 

%O = 100% - %C - %H = 100% - 40,00% - 6,67% = 53,33% 

Ta có: 

%C = \dfrac{12.x.100\%}{60}=40\%6012.x.100%​=40%

%H = \dfrac{1.y.100\%}{60}=6,67\%601.y.100%​=6,67%

%O = \dfrac{16.z.100\%}{60}=53.33\%6016.z.100%​=53.33%

⇒ x = 2 ; y = 4 ; z = 2. 

Vậy công thức của hợp chất là C2H4O2. 

$a)$ $\genfrac{}{}{0ex}{}{4}{9}+\genfrac{}{}{0ex}{}{1}{4}=\genfrac{}{}{0ex}{}{16}{36}+\genfrac{}{}{0ex}{}{9}{36}=\genfrac{}{}{0ex}{}{25}{36}$.

$b)$ $\genfrac{}{}{0ex}{}{1}{3}.\left(\genfrac{}{}{0ex}{}{-4}{5}\right)+\genfrac{}{}{0ex}{}{1}{3}.\genfrac{}{}{0ex}{}{-1}{5}+2023^0$

$=\genfrac{}{}{0ex}{}{1}{3}.(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{-1}{5})+1$

$=\genfrac{}{}{0ex}{}{1}{3}.(-1)+1$

$=-\genfrac{}{}{0ex}{}{1}{3}+1=\genfrac{}{}{0ex}{}{2}{3}$.

$c)$ $\genfrac{}{}{0ex}{}{1}{5}-[\genfrac{}{}{0ex}{}{1}{4}-(1-\genfrac{}{}{0ex}{}{1}{2}{)}^2]$.

$=\genfrac{}{}{0ex}{}{1}{5}-[\genfrac{}{}{0ex}{}{1}{4}-(\genfrac{}{}{0ex}{}{1}{2}{)}^2]$

$=\genfrac{}{}{0ex}{}{1}{5}-[\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{4}]$

$=\genfrac{}{}{0ex}{}{1}{5}-0=\genfrac{}{}{0ex}{}{1}{5}$
 

a) $x+\genfrac{}{}{0ex}{}{2}{5}=\genfrac{}{}{0ex}{}{-4}{3}$;

$x=\genfrac{}{}{0ex}{}{-4}{3}-\genfrac{}{}{0ex}{}{2}{5}$

$x=\genfrac{}{}{0ex}{}{-26}{15}$

b) $\genfrac{}{}{0ex}{}{-5}{6}+\genfrac{}{}{0ex}{}{1}{3}.x=(\genfrac{}{}{0ex}{}{-1}{2}{)}^2$;

$\genfrac{}{}{0ex}{}{-5}{6}+\genfrac{}{}{0ex}{}{1}{3}.x=\genfrac{}{}{0ex}{}{1}{4}$

$\genfrac{}{}{0ex}{}{1}{3}.x=\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{5}{6}$

$\genfrac{}{}{0ex}{}{1}{3}.x=\genfrac{}{}{0ex}{}{13}{12}$

$x=\genfrac{}{}{0ex}{}{13}{12}:\genfrac{}{}{0ex}{}{1}{3}$

$x=\genfrac{}{}{0ex}{}{13}{4}$

c) $\genfrac{}{}{0ex}{}{7}{12}-(x+\genfrac{}{}{0ex}{}{7}{6}).\genfrac{}{}{0ex}{}{6}{5}=(\genfrac{}{}{0ex}{}{-1}{2}{)}^3$.

$\genfrac{}{}{0ex}{}{7}{12}-(x+\genfrac{}{}{0ex}{}{7}{6}).\genfrac{}{}{0ex}{}{6}{5}=\genfrac{}{}{0ex}{}{-1}{8}$

$(x+\genfrac{}{}{0ex}{}{7}{6}).\genfrac{}{}{0ex}{}{6}{5}=\genfrac{}{}{0ex}{}{7}{12}-\left(\genfrac{}{}{0ex}{}{-1}{8}\right)$

$(x+\genfrac{}{}{0ex}{}{7}{6}).\genfrac{}{}{0ex}{}{6}{5}=\genfrac{}{}{0ex}{}{17}{24}$

$x+\genfrac{}{}{0ex}{}{7}{6}=\genfrac{}{}{0ex}{}{85}{144}$

$x=\genfrac{}{}{0ex}{}{85}{144}-\genfrac{}{}{0ex}{}{7}{6}$

$x=\genfrac{}{}{0ex}{}{-83}{144}$
 

a) $\widehat{mOx}+\widehat{xOn}=180^{\circ }$

Vậy $\widehat{nOx}=180^{\circ }-30^{\circ }=150^{\circ }$.

$Ot$ là tia phân giác của $\widehat{nOx}$, suy ra $\widehat{nOt}=\genfrac{}{}{0ex}{}{1}{2}.\widehat{nOx}=75^{\circ }$.

b) a // b suy ra $\widehat{A^4}=\widehat{B^2}=65^{\circ }$ (hai góc so le trong).

Mặt khác, ta có $\widehat{B^2}+\widehat{B^3}=180^{\circ }$

Suy ra $\widehat{B^3}=180^{\circ }-\widehat{B^2}=115^{\circ }$.