a: ta có: m⊥d

n⊥d

Do đó: m//n

b: Ta có: m//n

=>\(\hat{A_3}+\hat{B_1}=180^0\) (hai góc trong cùng phía)

=>\(\frac12\cdot\hat{B_1}+\hat{B_1}=180^0\)

=>\(\frac32\cdot\hat{B_1}=180^0\)

=>\(\hat{B_1}=180^0:\frac32=120^0\)

=>\(\hat{A_3}=120^0\cdot\frac12=60^0\)

Ta có: \(\hat{B_1}+\hat{B_2}=180^0\) (hai góc kề bù)

=>\(\hat{B_2}=180^0-120^0=60^0\)

c: Qua E, kẻ tia EF nằm giữa hai tia EA và EC sao cho EF//Am//Cn

Ta có: EF//Am

=>\(\hat{AEF}=\hat{EAm}=60^0\)

Ta có: EF//CB

=>\(\hat{FEC}=\hat{ECB}=80^0\)

Ta có: tia EF nằm giữa hai tia EA và EC

=>\(\hat{AEC}=\hat{AEF}+\hat{CEF}=60^0+80^0=140^0\)

Đặt \(\frac{a_1}{a_2}=\frac{a_2}{a_3}=\ldots=\frac{a_{2018}}{a_{2019}}=k\)

=>\(a_1=a_2\cdot k;a_2=a_3\cdot k;\ldots;a_{2018}=a_{2019}\cdot k\)

=>\(a_{2017}=a_{2019}\cdot k\cdot k=a_{2019}\cdot k^2\)

=>\(a_{2016}=a_{2019}\cdot k^2\cdot k=a_{2019}\cdot k^3\)

...

=>\(a_1=a_{2019}\cdot k^{2018}\)

\(\frac{a_1+a_2+\cdots+a_{2018}}{a_2+a_3+\cdots+a_{2019}}\)

\(=\frac{a_2\cdot k+a_3\cdot k+\cdots+a_{2019}\cdot k}{a_2+a_3+\cdots+a_{2019}}=k\)

=>\(\left(\frac{a_1+a_2+\cdots+a_{2018}}{a_2+a_3+\cdots+a_{2019}}\right)^{2018}=k^{2018}\) (1)

\(\frac{a_1}{a_{2019}}=\frac{a_{2019}\cdot k^{2018}}{a_{2019}}=k^{2018}\)

Do đó: \(\left(\frac{a_1+a_2+\cdots+a_{2018}}{a_2+a_3+\cdots+a_{2019}}\right)^{2018}=\frac{a_1}{a_{2019}}\)

a: Ta có: tia CA nằm giữa hai tia CB và CD

=>\(\hat{BCD}=\hat{BCA}+\hat{DCA}=80^0+30^0=110^0\)

ta có: \(\hat{BCD}+\hat{CBA}=110^0+70^0=180^0\)

mà hai góc này là hai góc ở vị trí trong cùng phía

nên AB//CD

b: AB//CD

=>\(\hat{BAC}=\hat{ACD}\) (hai góc so le trong)

=>\(\hat{BAC}=80^0\)

a, ta có A= 180 độ -70 độ -30 độ = 80 độ ( tổng 3 góc trong 1 tam giác = 180 độ )

mà AB=CD=80 độ nên AB//CD ( vì song song nên bằng nhau ) 1

b, góc BAC = 80 độ (1)

d: \(\frac27-\left(\frac23+2x\right)=\frac57\)

=>\(2x+\frac23=\frac27-\frac57=-\frac37\)

=>\(2x=-\frac37-\frac23=-\frac{9}{21}-\frac{14}{21}=-\frac{23}{21}\)

=>\(x=-\frac{23}{21}:2=-\frac{23}{42}\)

e: \(\frac12-2x=\left(-\frac12\right)^3\)

=>\(\frac12-2x=-\frac18\)

=>\(2x=\frac12+\frac18=\frac58\)

=>\(x=\frac58:2=\frac{5}{16}\)

f: \(\left(2x-3\right)\left(\frac34x+1\right)=0\)

=>\(\left[\begin{array}{l}2x-3=0\\ \frac34x+1=0\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=3\\ \frac34x=-1\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac32\\ x=-\frac43\end{array}\right.\)

g: \(\frac{7}{12}-\left(x+\frac76\right):\frac65=-\frac54\)

=>\(\left(x+\frac76\right):\frac65=\frac{7}{12}+\frac54=\frac{7}{12}+\frac{15}{12}=\frac{22}{12}=\frac{11}{6}\)

=>\(x+\frac76=\frac{11}{6}\cdot\frac65=\frac{11}{5}\)

=>\(x=\frac{11}{5}-\frac76=\frac{66}{30}-\frac{35}{30}=\frac{31}{30}\)

h: \(\frac34:\left(x+\frac12\right)-\frac56=-\frac14\)

=>\(\frac34:\left(x+\frac12\right)=-\frac14+\frac56=-\frac{3}{12}+\frac{10}{12}=\frac{7}{12}\)

=>\(x+\frac12=\frac34:\frac{7}{12}=\frac34\cdot\frac{12}{7}=\frac{36}{28}=\frac97\)

=>\(x=\frac97-\frac12=\frac{18}{14}-\frac{7}{14}=\frac{11}{14}\)

i: \(\frac25x+\frac35x=\frac34\)

=>\(x\left(\frac25+\frac35\right)=\frac34\)

=>\(x\cdot\frac55=\frac34\)

=>\(x=\frac34\)

k: \(\frac12x+\frac23x-x=\frac13\)

=>\(x\left(\frac12+\frac23-1\right)=\frac13\)

=>\(x\left(\frac12-\frac13\right)=\frac13\)

=>\(x\cdot\frac16=\frac13\)

=>\(x=\frac13:\frac16=2\)

l: \(\left(\frac32-\frac{2}{-5}\right):x-\frac12=\frac32\)

=>\(\left(\frac32+\frac25\right):x=\frac32+\frac12=2\)

=>\(\left(\frac{15}{10}+\frac{4}{10}\right):x=2\)

=>\(\frac{19}{10}:x=2\)

=>\(x=\frac{19}{10}:2=\frac{19}{20}\)

m: \(\left(5x-1\right)\left(2x-\frac13\right)=0\)

=>\(\left[\begin{array}{l}5x-1=0\\ 2x-\frac13=0\end{array}\right.\Rightarrow\left[\begin{array}{l}5x=1\\ 2x=\frac13\end{array}\right.\Rightarrow\left[\begin{array}{l}x=\frac15\\ x=\frac16\end{array}\right.\)

khó