a)

b)ta có:\(\widehat{ACB}\)=\(\widehat{CED}\)=30\(^O\)

Mà \(\widehat{ACB}\) và \(\widehat{CED}\) là 2 góc so le trong

Nên BC//DE

c)Xét△ABC có \(\widehat{A}\)+\(\widehat{B}\)+\(\widehat{C}\)=180\(^O\)

Hay​\(\widehat{A}\)+60\(^O\)+30​\(^O\)=180\(^O\)

    \(\widehat{A}\)=110\(^O\)

Vì \(\widehat{BAC}\) và \(\widehat{BAE}\) là 2 góc kề bù

Nên \(\widehat{BAC}\)+\(\widehat{BAE}\)=180\(^o\)

Hay 110\(^o\)+\(\widehat{BAE}\)=180\(^o\)

          \(\widehat{BAE}\)=90\(^o\)

 d

(1,41423)³:x=(1,41423)²

x=(1,41423)³:(1,41423)²

x=1,41423

) 0,2+137−65=0,2+173​−56​=

=15+107−65=51​+710​−56​

=15−65+107=51​−56​+710​

=−1+107=−1+710​

=37.=73​.

b) (45−1):35−23.0,5=(54​−1):53​−32​.0,5=

=−15:35−13=−51​:53​−31​

=−13−13=−31​−31​

=−23.=−32​.

) $xy//x^{\prime }{y}^{\prime }$ nên $\widehat{xAB}=\widehat{AB{y}^{\prime }}$ (hai góc so le trong). (1)

$A{A}^{\prime }$ là tia phân giác của $\widehat{xAB}$ nên: $\widehat{A^1}=\widehat{A^2}=\genfrac{}{}{0ex}{}{1}{2}\widehat{xAB}$. (2)

$B{B}^{\prime }$ là tia phân giác của $\widehat{AB{y}^{\prime }}$ nên: $\widehat{B^1}=\widehat{B^2}=\genfrac{}{}{0ex}{}{1}{2}\widehat{AB{y}^{\prime }}$. (3)

Từ (2) và (3) ta có: $\widehat{A^2}=\widehat{B^1}.$

Mà hai góc ở vị trí so le trong, nên từ (1), (2), (3) ta có: $A{A}^{\prime }$  //  $B{B}^{\prime }$ (có 2 góc so le trong bằng nhau).

b) $xy//x^{\prime }{y}^{\prime }$ nên $\widehat{A^1}=\widehat{A{A}^{\prime }B}$ (hai góc so le trong).

$A{A}^{\prime }//B{B}^{\prime }$ nên $\widehat{A^1}=\widehat{A{B}^{\prime }B}$ (hai góc đồng vị).

Vậy $\widehat{A{A}^{\prime }B}=\widehat{A{B}^{\prime }B}$.

Suy ra ${\hat{O}}^2+{\hat{A}}^2=180^{\circ }$ (2 góc trong cùng phía).

Khi đó ${\hat{O}}^1=\widehat{AOB}-{\hat{O}}^2=\widehat{AOB}-\left(180^{\circ }-{\hat{A}}^2\right)=\widehat{AOB}+{\hat{A}}^2-180^{\circ }={\hat{B}}^1$

$\Rightarrow Ot$ // $By$ (vì có cặp góc so le trong bằng nhau). (2)

Từ (1) và (2) suy ra $Ax$ // $By$ (vì cùng song song với $Ot$ ).

Vậy $At$ // $Bz$.

a)

$a)$ $\genfrac{}{}{0ex}{}{7}{4}\cdot x-\genfrac{}{}{0ex}{}{3}{2}=-\genfrac{}{}{0ex}{}{4}{5}$

$\genfrac{}{}{0ex}{}{7}{4}\cdot x=-\genfrac{}{}{0ex}{}{4}{5}+\genfrac{}{}{0ex}{}{3}{2}$

$\genfrac{}{}{0ex}{}{7}{4}\cdot x=\genfrac{}{}{0ex}{}{7}{10}$

$x=\genfrac{}{}{0ex}{}{7}{10}:\genfrac{}{}{0ex}{}{7}{4}$

$x=\genfrac{}{}{0ex}{}{2}{5}$

$b)$ $(x-\genfrac{}{}{0ex}{}{1}{4}{)}^2=\genfrac{}{}{0ex}{}{5}{36}-(\genfrac{}{}{0ex}{}{1}{3}{)}^2$

$(x-\genfrac{}{}{0ex}{}{1}{4}{)}^2=\genfrac{}{}{0ex}{}{5}{36}-\genfrac{}{}{0ex}{}{1}{9}$

$(x-\genfrac{}{}{0ex}{}{1}{4}{)}^2=\genfrac{}{}{0ex}{}{1}{36}$

$x-\genfrac{}{}{0ex}{}{1}{4}=\genfrac{}{}{0ex}{}{1}{6}$ hoặc $x-\genfrac{}{}{0ex}{}{1}{4}=-\genfrac{}{}{0ex}{}{1}{6}$

$x=\genfrac{}{}{0ex}{}{5}{12}$ hoặc $x=\genfrac{}{}{0ex}{}{1}{12}$

$c)$ $-x+\genfrac{}{}{0ex}{}{3}{2}=x+\genfrac{}{}{0ex}{}{3}{5}$

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

$-2x=-\genfrac{}{}{0ex}{}{9}{10}$

$x=\genfrac{}{}{0ex}{}{9}{20}$
 

a) $A=\genfrac{}{}{0ex}{}{15}{12}+\genfrac{}{}{0ex}{}{5}{13}+\genfrac{}{}{0ex}{}{-3}{12}+\genfrac{}{}{0ex}{}{-18}{13}$

$=(\genfrac{}{}{0ex}{}{15}{12}+\genfrac{}{}{0ex}{}{-3}{12})+(\genfrac{}{}{0ex}{}{5}{13}+\genfrac{}{}{0ex}{}{-18}{13})$

$=\left(\genfrac{}{}{0ex}{}{12}{12}\right)+\left(\genfrac{}{}{0ex}{}{-13}{13}\right)$

$=1+(-1)=0$

b) $B=\genfrac{}{}{0ex}{}{11}{15}\cdot \genfrac{}{}{0ex}{}{-19}{13}+\genfrac{}{}{0ex}{}{-7}{13}\cdot \genfrac{}{}{0ex}{}{11}{15}$

$=\genfrac{}{}{0ex}{}{11}{15}\cdot (\genfrac{}{}{0ex}{}{-19}{13}+\genfrac{}{}{0ex}{}{-7}{13})$

$=\genfrac{}{}{0ex}{}{11}{15}\cdot (-2)$

$=\genfrac{}{}{0ex}{}{-22}{15}$

c) $C=2022^0-(\genfrac{}{}{0ex}{}{1}{7}{)}^5\cdot 7^5$

$=1-\genfrac{}{}{0ex}{}{1}{7^5}\cdot 7^5$

$=1-1=0$