a) $\mathrm{A}=\left[\genfrac{}{}{0ex}{}{2}{7}\left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{3}\right)\right]:\left[\genfrac{}{}{0ex}{}{2}{7}\left(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{2}{5}\right)\right]=\left(\genfrac{}{}{0ex}{}{1}{4}-\genfrac{}{}{0ex}{}{1}{3}\right):\left(\genfrac{}{}{0ex}{}{1}{3}-\genfrac{}{}{0ex}{}{2}{5}\right)=1\genfrac{}{}{0ex}{}{1}{4}$.
b) $\mathrm{B}=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{3}{4}\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{2}{7}-\genfrac{}{}{0ex}{}{1}{3}+\genfrac{}{}{0ex}{}{2}{7}\right)}{\genfrac{}{}{0ex}{}{1}{5}\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)-\genfrac{}{}{0ex}{}{1}{3}\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)}=\genfrac{}{}{0ex}{}{\genfrac{}{}{0ex}{}{3}{4}\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{3}\right)}{\left(\genfrac{}{}{0ex}{}{1}{5}-\genfrac{}{}{0ex}{}{1}{3}\right)\left(\genfrac{}{}{0ex}{}{2}{7}+\genfrac{}{}{0ex}{}{1}{3}\right)}=1\genfrac{}{}{0ex}{}{11}{52}$.

a) $\mathrm{A}=\genfrac{}{}{0ex}{}{3}{5}.\genfrac{}{}{0ex}{}{6}{7}+\genfrac{}{}{0ex}{}{3}{7}.\genfrac{}{}{0ex}{}{3}{5}-\genfrac{}{}{0ex}{}{2}{7}.\genfrac{}{}{0ex}{}{3}{5}$
$=\genfrac{}{}{0ex}{}{3}{5}\cdot \left(\genfrac{}{}{0ex}{}{6}{7}+\genfrac{}{}{0ex}{}{3}{7}-\genfrac{}{}{0ex}{}{2}{7}\right)=\genfrac{}{}{0ex}{}{3}{5}$
b) $\mathrm{B}=\left(-13\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{-2}{9}\cdot \genfrac{}{}{0ex}{}{2}{5}+\genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{5}{2}$
$=\left(-13-\genfrac{}{}{0ex}{}{2}{9}+\genfrac{}{}{0ex}{}{11}{9}\right)\cdot \genfrac{}{}{0ex}{}{2}{5}\cdot \genfrac{}{}{0ex}{}{5}{2}=-13+\left(\genfrac{}{}{0ex}{}{11}{9}-\genfrac{}{}{0ex}{}{2}{9}\right)=-12.$
c) $\mathrm{C}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}+\left(\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{-1}{5}+\genfrac{}{}{0ex}{}{2}{7}\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=\left(\left(\genfrac{}{}{0ex}{}{-4}{5}+\genfrac{}{}{0ex}{}{-1}{5}\right)+\left(\genfrac{}{}{0ex}{}{5}{7}+\genfrac{}{}{0ex}{}{2}{7}\right)\right)\cdot \genfrac{}{}{0ex}{}{3}{2}=0.$
d) $\mathrm{D}=\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{1}{15}-\genfrac{}{}{0ex}{}{10}{15}\right)+\genfrac{}{}{0ex}{}{4}{9}:\left(\genfrac{}{}{0ex}{}{2}{22}-\genfrac{}{}{0ex}{}{5}{22}\right)$
$=\genfrac{}{}{0ex}{}{4}{9}:\genfrac{}{}{0ex}{}{-3}{5}+\genfrac{}{}{0ex}{}{4}{9}:\genfrac{}{}{0ex}{}{-3}{22}=\genfrac{}{}{0ex}{}{4}{9}\cdot \genfrac{}{}{0ex}{}{-5}{3}+\genfrac{}{}{0ex}{}{4}{9}.\genfrac{}{}{0ex}{}{-22}{3}$
$=\genfrac{}{}{0ex}{}{4}{9}\cdot \left(\genfrac{}{}{0ex}{}{-5}{3}+\genfrac{}{}{0ex}{}{-22}{3}\right)=\genfrac{}{}{0ex}{}{4}{9}.\genfrac{}{}{0ex}{}{-27}{3}=-4.$

a) $P=\genfrac{}{}{0ex}{}{2}{3}+\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{3}{5}-\genfrac{}{}{0ex}{}{7}{45}+\genfrac{}{}{0ex}{}{5}{9}+\genfrac{}{}{0ex}{}{1}{12}+\genfrac{}{}{0ex}{}{1}{35}$ $=\left(\genfrac{}{}{0ex}{}{2}{3}+\genfrac{}{}{0ex}{}{1}{4}+\genfrac{}{}{0ex}{}{1}{12}\right)+\left(\genfrac{}{}{0ex}{}{5}{9}-\genfrac{}{}{0ex}{}{7}{45}\right)+\genfrac{}{}{0ex}{}{3}{5}+\genfrac{}{}{0ex}{}{1}{35}=1+\genfrac{}{}{0ex}{}{4}{5}+\genfrac{}{}{0ex}{}{3}{5}+\genfrac{}{}{0ex}{}{1}{35}=2\genfrac{}{}{0ex}{}{1}{35}$.

(Chú ý: ta gói các số hạng có mẫu dễ quy đồng hơn với nhau.)
b) $Q=(5-6-2)+\left(-\genfrac{}{}{0ex}{}{3}{4}-\genfrac{}{}{0ex}{}{7}{4}+\genfrac{}{}{0ex}{}{5}{4}\right)+\left(\genfrac{}{}{0ex}{}{1}{5}+\genfrac{}{}{0ex}{}{8}{5}-\genfrac{}{}{0ex}{}{16}{5}\right)=-\left(3+\genfrac{}{}{0ex}{}{5}{4}+\genfrac{}{}{0ex}{}{7}{5}\right)$ $=-\genfrac{}{}{0ex}{}{113}{20}$.

a) $A=\left(\genfrac{}{}{0ex}{}{1}{3}+\genfrac{}{}{0ex}{}{2}{3}\right)-\left(\genfrac{}{}{0ex}{}{8}{15}+\genfrac{}{}{0ex}{}{7}{15}\right)+\left(\genfrac{}{}{0ex}{}{-1}{7}+1\genfrac{}{}{0ex}{}{1}{7}\right)=1-1+1=1$;
b) $\mathrm{B}=\left(0.25-1\genfrac{}{}{0ex}{}{1}{4}\right)+\left(\genfrac{}{}{0ex}{}{3}{5}+\genfrac{}{}{0ex}{}{2}{5}\right)-\genfrac{}{}{0ex}{}{1}{8}$
$=\left(\genfrac{}{}{0ex}{}{1}{4}-1-\genfrac{}{}{0ex}{}{1}{4}\right)+1-\genfrac{}{}{0ex}{}{1}{8}=\genfrac{}{}{0ex}{}{-1}{8}$.

Ta có $\widehat{zOy}=\widehat{xOy}+\widehat{yOz}=4\cdot \widehat{yOz}+\widehat{yOz}=5\cdot \widehat{yOz}$ (1).

Mà $\widehat{yOt}=90^{\circ }\iff 90^{\circ }=\widehat{yOz}+\widehat{zOt}=\widehat{yOz}+\genfrac{}{}{0ex}{}{1}{2}\widehat{xOz}=3.\widehat{yOz}\iff \widehat{yOz}=30^{\circ }$ (2) .

Thay (2) vào (1), ta được: $xOz=5.30^{\circ }=150^{\circ }$.

Vậy $\widehat{xOy}=150^{\circ }$.

Vì các tia $OC$ và $OD$ ở trong góc $\widehat{AOB}$ nên:

$\widehat{AOD}=\widehat{AOC}-\widehat{COD}=90^{\circ }-\widehat{COD}$ (1)

$\widehat{BOC}=\widehat{BOD}-\widehat{COD}=90^{\circ }-\widehat{COD}$ (2)

Từ (1) và (2), suy ra: $\widehat{AOD}=\widehat{BOC}$.

b) Ta có

$\widehat{AOB}+\widehat{COD}=(\widehat{AOC}+\widehat{BOC})+\widehat{COD}=\widehat{AOC}+\widehat{BOC}+\widehat{COD}=\widehat{AOC}+\widehat{BOD}=90^{\circ }+90^{\circ }=180^{\circ }$

c) Từ giả thiết, ta có: $\widehat{AOD}=2\cdot \widehat{xOD}$.

Mà $\widehat{xOy}=\widehat{xOD}+\widehat{DOC}+\widehat{COy}=2\cdot \widehat{xOD}+\widehat{DOC}=\widehat{AOD}+\widehat{DOC}=\widehat{AOC}=90^{\circ }$.

Vậy $Ox\perp Oy$.

Từ hình vẽ ta thấy:

$\widehat{O^1}=\widehat{O^3}$ (hai góc đối đỉnh);

$\widehat{O^2}=\widehat{O^4}$ (hai góc đối đỉnh).

Mà $Ot$ là tia phân giác của góc $xOy$ nên $\widehat{O^1}=\widehat{O^2}$.

Suy ra $\widehat{O^3}=\widehat{O^4}$.

Mà tia $O{t}^{\prime }$ nằm giữa hai tia $O{x}^{\prime }$ và $O{y}^{\prime }$ nên $O{t}^{\prime }$ là tia phân giác của góc $x^{\prime }O{y}^{\prime }$.

Xét góc $xOy$ có góc kề bù là góc $xOz$.

Gọi tia $Ot$, $Ok$ lần lượt là tia phân giác của góc $xOy$ và góc $xOz$.

Khi đó, ta có:

$180^{\circ }=\widehat{xOy}+\widehat{xOz}=2.\widehat{xOt}+2.\widehat{xOk}$

Suy ra $\widehat{xOt}+\widehat{xOk}=90^{\circ }$.

Vậy $Ot\perp Ok$.

 Biết $\widehat{O^1}-\widehat{O^2}=70^{\circ }$

Suy ra $\widehat{O^1}=\widehat{O^2}+70^{\circ }$

Mà $\widehat{O^1}$ và $\widehat{O^2}$ là hai góc kề bù nên $\widehat{O^1}+\widehat{O^2}=180^{\circ }$.

Thay $\widehat{O^1}=\widehat{O^2}+70^{\circ }$ ta được $\widehat{O^2}+\widehat{O^2}+70^{\circ }=180^{\circ }$

Hay $2.\widehat{O^2}=110^{\circ }$

Suy ra $\widehat{O^2}=55^{\circ }$.

Mà hai góc $\widehat{O^2}$ và $\widehat{O^4}$ đối đỉnh nên $\widehat{O^4}=55^{\circ }$

 Biết $\widehat{O^1}+\widehat{O^2}+\widehat{O^3}=325^{\circ }$.

Mà $\widehat{O^1}$ và $\widehat{O^2}$ là hai góc kề bù nên $\widehat{O^1}+\widehat{O^2}=180^{\circ }$.

Suy ra $\widehat{O^3}=325^{\circ }-180^{\circ }=145^{\circ }$.

Mà $\widehat{O^3}$ và $\widehat{O^4}$ là hai góc kề bù nên $\widehat{O^4}=180^{\circ }-145^{\circ }=35^{\circ }$.