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

Bài 3:

a: \(A=3^2\cdot\frac{1}{243}\cdot81^2\cdot\frac{1}{3^3}\)

\(=\frac{9}{243}\cdot81\cdot81\cdot\frac{1}{27}\)

\(=\frac{1}{27}\cdot81\cdot3=3\cdot3=9\)

b: \(B=\left(4\cdot2^5\right):\left(2^3\cdot\frac{1}{16}\right)\)

\(=2^2\cdot2^5:\left(\frac{2^3}{16}\right)=2^7:\frac12=2^7\cdot2=2^8=256\)

Bài 2:

a: \(A=\left(3^2\right)^2-\left(-2^3\right)^2-\left(-5^2\right)^2\)

\(=3^4-2^6-\left(-25\right)^2\)

=81-64-625

=17-625

=-608

b: \(B=2^3+3\cdot\left(\frac12\right)^0\cdot\left(\frac12\right)^2\cdot4+\left\lbrack\left(-2\right)^2:\frac12\right\rbrack:8\)

\(=8+3\cdot1\cdot\frac14\cdot4+4\cdot\frac28\)

=8+3+1

=11+1

=12

Bài 1:

a: \(\left(\frac23\right)^3\cdot\left(-\frac34\right)^2\cdot\left(-1\right)^5:\left(\frac25\right)^2\cdot\left(-\frac{5}{12}\right)^2\)

\(=\frac{2^3}{3^3}\cdot\frac{3^2}{4^2}\cdot\left(-1\right):\frac{4}{25}\cdot\frac{25}{144}\)

\(=\frac{2^3}{2^4}\cdot\frac13\cdot\left(-1\right)\cdot\frac{25}{4}\cdot\frac{25}{144}=\frac16\cdot\left(-1\right)\cdot\frac{625}{576}=\frac{-625}{3456}\)

b:Sửa đề: \(\frac{\left(6^6+6^3\cdot3^3+3^6\right)}{-73}\)

\(=\frac{3^6\cdot2^6+3^6\cdot2^3+3^6}{-73}\)

\(=\frac{3^6\left(2^6+2^3+1\right)}{-73}=\frac{3^6\cdot73}{-73}=-3^6=-729\)

Bài 4:

Ta có: \(\hat{M_2}=\hat{N_2}\left(=60^0\right)\)

mà hai góc này là hai góc ở vị trí đồng vị

nên a//b

Bài 3:

a//b

a⊥BA

Do đó: b⊥BA

=>\(\hat{ABC}=90^0\)

AD//BC

=>\(\hat{ADC}+\hat{DCB}=180^0\)

=>\(\hat{ADC}=180^0-110^0=70^0\)

Bài 2:

a: \(-\frac35+\frac{-2}{5}:x=\frac13\)

=>\(-\frac25:x=\frac13+\frac35=\frac{5}{15}+\frac{9}{15}=\frac{14}{15}\)

=>\(x=-\frac25:\frac{14}{15}=-\frac25\cdot\frac{15}{14}=-\frac37\)

b: \(0,2+\left|x-1,3\right|=1,5\)

=>|x-1,3|=1,5-0,2=1,3

=>\(\left[\begin{array}{l}x-1,3=1,3\\ x-1,3=-1,3\end{array}\right.\Rightarrow\left[\begin{array}{l}x=2,6\\ x=0\end{array}\right.\)

c: \(\left(\frac37-2x\right)^2=\frac49\)

=>\(\left[\begin{array}{l}\frac37-2x=\frac23\\ \frac37-2x=-\frac23\end{array}\right.\Rightarrow\left[\begin{array}{l}2x=\frac37-\frac23=\frac{9}{21}-\frac{14}{21}=-\frac{5}{21}\\ 2x=\frac37+\frac23=\frac{9}{21}+\frac{14}{21}=\frac{23}{21}\end{array}\right.\)

=>\(\left[\begin{array}{l}x=-\frac{5}{21}:2=-\frac{5}{42}\\ x=\frac{23}{21}:2=\frac{23}{42}\end{array}\right.\)

d: \(2^{x}+2^{x+3}=144\)

=>\(2^{x}+2^{x}\cdot2^3=144\)

=>\(2^{x}\left(1+2^3\right)=144\)

=>\(2^{x}\cdot9=144\)

=>\(2^{x}=\frac{144}{9}=16=2^4\)

=>x=4

Bài 1:

a: \(\frac{14}{57}+\frac{29}{23}-\frac{71}{57}+\frac{-6}{23}\)

\(=\left(\frac{14}{57}-\frac{71}{57}\right)+\left(\frac{29}{23}-\frac{6}{23}\right)\)

\(=\frac{-57}{57}+\frac{23}{23}=-1+1=0\)

b: \(\frac{5}{12}\cdot\left(-\frac34\right)+\frac{7}{12}\left(-\frac34\right)\)

\(=-\frac34\left(\frac{5}{12}+\frac{7}{12}\right)=-\frac34\cdot\frac{12}{12}=-\frac34\)

d: \(\left(-\frac{3}{11}:\frac{5}{22}\right)\cdot\left(-\frac{15}{3}:\frac{26}{3}\right)\)

\(=-\frac{3}{11}\cdot\frac{22}{5}\cdot\left(_{}-5\right)\cdot\frac{3}{26}=-\frac35\cdot\left(-5\right)\cdot2\cdot\frac{3}{26}=3\cdot2\cdot\frac{3}{26}=\frac{9}{13}\)

f: \(\frac{9^{15}\cdot8^{11}}{3^{29}\cdot16^8}=\frac{3^{30}}{3^{29}}\cdot\frac{2^{33}}{2^{32}}=3\cdot2=6\)