\(\text{Áp dụng t/c tỉ lệ thức ta có thể b/đ như sau:}\)

\(\frac{x+1}{x+2}=\frac{2}{3}\Rightarrow\left(x+1\right)3=\left(x+2\right)2\)

\(\Rightarrow3x+3=2x+4\)

\(\Rightarrow2x-3x=3-4\)

\(\Rightarrow-x=-1\)

\(\Rightarrow x=1\)

 \(\frac{329}{1051}=\frac{1}{3+\frac{1}{5+\frac{1}{a+\frac{1}{b}}}}\)

Ta có: \(\frac{329}{1051}=\frac{1}{\frac{1051}{329}}\rightarrow\frac{1}{3+\frac{1}{5+\frac{1}{a+\frac{1}{b}}}}=\frac{1}{\frac{1051}{329}}\rightarrow3+\frac{1}{3+\frac{1}{5+\frac{1}{a+\frac{1}{b}}}}=\frac{1051}{329}\)

  mà \(\frac{1051}{329}=3+\frac{64}{329}=3+\frac{1}{\frac{329}{64}}\rightarrow\frac{1}{5+\frac{1}{a+\frac{1}{b}}}\rightarrow5+\frac{1}{a+\frac{1}{b}}=\frac{329}{64}\)

Tương tự: \(\frac{329}{64}=5+\frac{9}{64}=5+\frac{1}{\frac{64}{9}}\rightarrow\frac{1}{a+\frac{1}{b}}=\frac{1}{\frac{64}{9}}\rightarrow a+\frac{1}{b}=\frac{64}{9}\)

mà \(\frac{64}{9}=7+\frac{1}{9}\Leftrightarrow\frac{64}{9}=a+\frac{1}{b}\Rightarrow a=7,b=9\)

Vậy \(a=7,b=9\)

\(\frac{329}{1051}=\frac{1}{3+\frac{1}{5+\frac{1}{7+\frac{1}{9}}}}\)\(\Rightarrow a=7;b=9\)

Vi \(\left(\frac{1}{7}x-\frac{2}{7}\right)\cdot\left(-\frac{1}{5}x+\frac{3}{5}\right)\cdot\left(\frac{1}{3}x+\frac{4}{3}\right)=0\)

\(\Rightarrow\hept{\begin{cases}\frac{1}{7}x-\frac{2}{7}=0\\-\frac{1}{5}x+\frac{3}{5}=0\\\frac{1}{3}x+\frac{4}{3}=0\end{cases}\Rightarrow\hept{\begin{cases}\frac{1}{7}x=\frac{2}{7}\\-\frac{1}{5}x=-\frac{3}{5}\\\frac{1}{3}x=-\frac{4}{3}\end{cases}\Rightarrow}\hept{\begin{cases}x=2\\x=3\\x=-4\end{cases}}}\)

Vậy \(x\in\left\{-4;3;2\right\}\)

\(\Rightarrow\frac{1}{7}x-\frac{2}{7}=0\text{ hoặc }-\frac{1}{5}x+\frac{3}{5}=0\text{ hoặc }\frac{1}{3}x+\frac{4}{3}=0\)

\(\Rightarrow x=2\text{ hoặc }x=3\text{ hoặc }x=-4\)

Vậy tập nghiệm của pt là \(S=\left\{2;3;-4\right\}\) 

a, 5^n +5^(n+2)

=5(5^n-1+5ⁿ⁺¹)

=26*5ⁿ

b,2/3. 3^(n-1)

=2*3^n-2