Bài làm

Vì \(\left(\frac{1}{5}-\frac{3}{2}.x\right)^2=\frac{9}{4}\)

\(\Rightarrow\left(\frac{1}{5}-\frac{3}{2}.x\right)^2=\left(\frac{3}{2}\right)^2\)

\(\Rightarrow\frac{1}{5}-\frac{3}{2}x=\frac{3}{2}\)

\(\Rightarrow\frac{3}{2}x=\frac{1}{5}-\frac{3}{2}\)

\(\Rightarrow\frac{3}{2}x=\frac{2}{10}-\frac{15}{10}\)

\(\Rightarrow\frac{3}{2}x=-\frac{13}{10}\)

\(\Rightarrow x=-\frac{13}{10}.\frac{2}{3}\)

\(\Rightarrow x=-\frac{13}{15}\)

Vậy \(x=-\frac{13}{15}\)

\(\left(\frac{1}{5}-\frac{3}{2}\cdot x\right)^2=\frac{9}{4}\)

\(\orbr{\begin{cases}\frac{1}{5}-\frac{3}{2}\cdot x=\frac{3}{2}\\\frac{1}{5}-\frac{3}{2}\cdot x=-\frac{3}{2}\end{cases}}\)

\(\orbr{\begin{cases}\frac{3}{2}\cdot x=\frac{1}{5}-\frac{3}{2}=-\frac{13}{10}\\\frac{3}{2}\cdot x=\frac{1}{5}-\left(-\frac{3}{2}\right)=\frac{1}{5}+\frac{3}{2}=\frac{17}{10}\end{cases}}\)

\(\orbr{\begin{cases}x=-\frac{13}{10}:\frac{3}{2}=-\frac{13}{10}\cdot\frac{2}{3}=-\frac{13}{15}\\x=\frac{17}{10}:\frac{3}{2}=\frac{17}{10}\cdot\frac{2}{3}=\frac{17}{15}\end{cases}}\)

Vậy \(x=-\frac{13}{15}\)hoặc \(x=\frac{17}{15}\)

bn cứ kiếm trên mạng ấy, nhiều lắm 

\(3^{-1}\cdot3^n+4\cdot3^n=13\cdot3^5\)

=> \(\frac{1}{3}\cdot3^n+4\cdot3^n=13\cdot3^5\)

=> \(3^n\left[\frac{1}{3}+4\right]=13\cdot3^5\)

=> \(3^n\cdot\frac{13}{3}=13\cdot3^5\)

=> \(3^n=\frac{13\cdot3^5}{\frac{13}{3}}=729\)

=> \(3^n=3^6\)

=> n = 6

Vậy giá trị của n cần tìm là 6

mình mới lớp 6 mẹ bắt làm rồi :)))

\(A=1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^4-...-\left(\frac{3}{4}\right)^{2009}+\left(\frac{3}{4}\right)^{2010}\)

\(\Rightarrow\frac{3}{4}A=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^2+-\left(\frac{3}{4}\right)^4+...+\left(\frac{3}{4}\right)^{2010}-\left(\frac{3}{4}\right)^{2011}\)

\(\Rightarrow\frac{3}{4}A+A=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^2+-\left(\frac{3}{4}\right)^4+...+\left(\frac{3}{4}\right)^{20010}-\left(\frac{3}{4}\right)^{2011}\)

\(+1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^4-...-\left(\frac{3}{4}\right)^{2009}+\left(\frac{3}{4}\right)^{2010}\)

\(\Rightarrow\frac{7}{4}A=1-\left(\frac{3}{4}\right)^{2011}\)

\(\Rightarrow A=\frac{4}{7}-\frac{4}{7}.\left(\frac{3}{4}\right)^{2011}\)

\(\Rightarrow A=\frac{4}{7}-\frac{3^{2011}}{7.4^{2010}}\)

Vậy A không là số tự nhiên

Số nguyên chứ không pk stn nhé, nhầm