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: \(11^{x-1}=11^7\)

=>x-1=7

=>x=7+1=8

b: \(\left(x-4\right)^2=64\)

=>\(\left[\begin{array}{l}x-4=8\\ x-4=-8\end{array}\right.\Rightarrow\left[\begin{array}{l}x=8+4=12\\ x=-8+4=-4\end{array}\right.\)

c: \(5^{x+1}-5^{x}=100\cdot25^{29}\)

=>\(5^{x}\cdot5-5^{x}=4\cdot5^2\cdot5^{29}=4\cdot5^{31}\)

=>\(5^{x}\cdot4=4\cdot5^{31}\)

=>x=31

your gay

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