Kẻ tia $Cx$ là tia phân giác của $\widehat{ACD}$ và $Dy$ là tia phân giác của $\widehat{BDC}$, hai tia $Cx$ và $Dy$ cắt nhau tại $E$.

$\widehat{C^1}=\widehat{C^2}=60^{\circ }$ và $\widehat{D^1}=\widehat{D^2}=30^{\circ }$

Kẻ tia $Ez//m//n$, tính $\widehat{E^1}=60^{\circ }$ và $\widehat{E^2}=30^{\circ }$

Suy ra $\widehat{CED}=90^{\circ }$.

a) Ta có: {�4^=110∘�2^=110∘ ⇒�4^=�2^=110∘{​A4​​=110∘B2​​=110∘​ ⇒A4​​=B2​​=110∘.

Mà hai góc ờ vị trí so le trong $\Rightarrow$ $a//b$.

b) Ta có: {� ⊥��//�⇒�⊥�{​c ⊥aa//b​⇒c⊥b

c) Vì $a//b\Rightarrow \widehat{A^4}+\widehat{B^1}=180^{\circ }$

Mà hai góc ở vị trí trong cùng phía $\Rightarrow \widehat{B^1}=180^{\circ }-\widehat{A^4}=70^{\circ }$.

Vì $b\perp c$; $e\perp c$ và $b//e$

$\Rightarrow \widehat{B^2}=\widehat{C^2}=110^{\circ }$ (hai góc ở vị trí đồng vị)

Ta có $\widehat{C^2}$ và $\widehat{C^3}$ là hai góc kề bù $\Rightarrow \widehat{C^2}+\widehat{C^3}=180^{\circ }$

$\Rightarrow \widehat{C^3}=180^{\circ }-\widehat{C^2}=70^{\circ }$.

+ Hai cặp góc so le trong: góc B1 và góc A3; góc A4 và góc B2 .

+ Bốn cặp góc đồng vị: góc B1 và góc A1; góc B2 và góc A2; góc B3 và góc A3; góc B4 và góc A4 .

a) $\genfrac{}{}{0ex}{}{2}{3}\cdot \genfrac{}{}{0ex}{}{5}{4}-\genfrac{}{}{0ex}{}{3}{4}\cdot \genfrac{}{}{0ex}{}{2}{3}=\genfrac{}{}{0ex}{}{2}{3}\cdot \left(\genfrac{}{}{0ex}{}{5}{4}-\genfrac{}{}{0ex}{}{3}{4}\right)=\genfrac{}{}{0ex}{}{2}{3}\cdot \genfrac{}{}{0ex}{}{1}{2}=\genfrac{}{}{0ex}{}{1}{3}$;
b) $2\cdot {\left(\genfrac{}{}{0ex}{}{-3}{2}\right)}^2-\genfrac{}{}{0ex}{}{7}{2}=2\cdot \genfrac{}{}{0ex}{}{9}{4}-\genfrac{}{}{0ex}{}{7}{2}=\genfrac{}{}{0ex}{}{9}{2}-\genfrac{}{}{0ex}{}{7}{2}=1$;
c) $-\genfrac{}{}{0ex}{}{3}{4}\cdot 5\genfrac{}{}{0ex}{}{3}{13}-0,75\cdot \genfrac{}{}{0ex}{}{36}{13}=-\genfrac{}{}{0ex}{}{3}{4}\cdot 5\genfrac{}{}{0ex}{}{3}{13}-\genfrac{}{}{0ex}{}{3}{4}\cdot \genfrac{}{}{0ex}{}{36}{13}$
$=-\genfrac{}{}{0ex}{}{3}{4}\left(5\genfrac{}{}{0ex}{}{3}{13}+\genfrac{}{}{0ex}{}{36}{13}\right)$
$=-\genfrac{}{}{0ex}{}{3}{4}\cdot 8=-6$

a) $\genfrac{}{}{0ex}{}{1}{5}+\genfrac{}{}{0ex}{}{4}{5}:x=0,75$

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

$\genfrac{}{}{0ex}{}{4}{5}:x=\genfrac{}{}{0ex}{}{3}{4}-\genfrac{}{}{0ex}{}{1}{5}$

$\genfrac{}{}{0ex}{}{4}{5}:x=\genfrac{}{}{0ex}{}{11}{20}$

$x=\genfrac{}{}{0ex}{}{16}{11}$;

b) $x+\genfrac{}{}{0ex}{}{1}{2}=1-x$

 $2x=1-\genfrac{}{}{0ex}{}{1}{2}$

 $2x=\genfrac{}{}{0ex}{}{1}{2}$

 $x=\genfrac{}{}{0ex}{}{1}{4}$.