a) x=2

b) x=6

a)\(3.\left(5^x-1\right)-2=70\\3.\left(5^x-1\right)=72\\5^x-1=24\\ 5^x=25=5^2 \\ \Rightarrow x=2 \)

b)\(2^x+2^{x+1}+2^{x+2}=960-2^{x+3}\\2^x+2^{x+1}+2^{x+2}+2^{x+3}=960\\ 2^x\left(1+2+4+8\right)=2^x.15=960\\ 2^x=64=2^6\\ x=6\)

\(\frac{11}{12}+\frac{7}{16}\div\frac{-3}{4}-\frac{13}{12}=\)\(\frac{11}{12}+\frac{-7}{12}-\frac{13}{12}=\)\(\frac{-3}{4}\)

\(\left(\frac{1}{2}-\frac{2}{3}\right)\times\frac{4}{5}\)

\(=\frac{-1}{6}\times\frac{4}{5}\)

\(=\frac{-2}{15}\)

\(\left(\frac{-4}{9}\right)\times\frac{3}{8}+\frac{1}{18}\)

\(=\frac{-1}{6}+\frac{1}{18}\)

\(=\frac{-1}{9}\)

\(\frac{2}{7}+\frac{28}{27}\div\left(\frac{-4}{9}\right)\)

\(=\frac{2}{7}+\frac{-7}{3}\)

\(=\frac{-43}{21}\)

bấm vào tìm kiếm sẽ có câu tương tự nha em

\(\frac{n}{n+1}\)và \(\frac{n+2016}{n+2017}\).

Ta có:

\(\frac{n}{n+1}=\frac{n+1-1}{n+1}=1-\frac{1}{n+1}\)

\(\frac{n+2016}{n+2017}=\frac{n+2017-1}{n+2017}=1-\frac{1}{n+2017}\).

Ta lại có:

\(\frac{1}{n+1}>\frac{1}{n+2017}\forall n\).

\(\Rightarrow\frac{-1}{n+1}< \frac{-1}{n+2017}\forall n\).

\(\Rightarrow1-\frac{1}{n+1}< 1-\frac{1}{n+2017}\forall n\).

Do đó \(\frac{n}{n+1}< \frac{n+2016}{n+2017}\).

Vậy....

Ta có : \(2A=2+1+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}\)

\(\Rightarrow2A-A=3+\frac{1}{2}+\frac{1}{2^2}+...+\frac{1}{2^{2016}}-1-\frac{1}{2}-\frac{1}{2^2}-\frac{1}{2^3}-...-\frac{1}{2^{2017}}\)

\(=2-\frac{1}{2^{2017}}=\frac{2^{2018}-1}{2^{2017}}\)

dunng ko vay