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

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

\(\frac{3}{4}A-1=-\left[1-\frac{3}{4}+\left(\frac{3}{4}\right)^2-\left(\frac{3}{4}\right)^3+...+\left(\frac{3}{4}\right)^{2010}\right]-\left(\frac{3}{4}\right)^{2011}\)

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

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

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

\(A=\frac{1-\left(\frac{3}{4}\right)^{2011}}{-\frac{1}{4}}=1:-\frac{1}{4}-\left(\frac{3}{4}\right)^{2011}:\left(-\frac{1}{4}\right)=-4+3\cdot\left(\frac{3}{4}\right)^{2010}\)

=>A không phải là số nguyên

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

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

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

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

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

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

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

Vậy A không là số tự nhiên

Số nguyên chứ không pk stn nhé, nhầm

Sửa bài:

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

=> \(\frac{3}{4}.A=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\left(\frac{3}{4}\right)^4+\left(\frac{3}{4}\right)^5-...-\left(\frac{3}{4}\right)^{2010}+\left(\frac{3}{4}\right)^{2011}\)

=> \(A+\frac{3}{4}A=1+\left(\frac{3}{4}\right)^{2011}\)

<=> \(\frac{7}{4}A=1+\left(\frac{3}{4}\right)^{2011}\)

=> \(A=\frac{4}{7}+\frac{4}{7}.\left(\frac{3}{4}\right)^{2011}\)không phải là số nguyên .

tich di mk giai cho

\(\frac{3}{4}A=\frac{3}{4}-\left(\frac{3}{4}\right)^2+\left(\frac{3}{4}\right)^3-\left(\frac{3}{4}\right)^4+\left(\frac{3}{4}\right)^5-....-\left(\frac{3}{4}\right)^{2010}\)

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

\(\frac{7}{4}A=1-\left(\frac{3}{4}\right)^{2010}\)

\(A=\frac{4}{7}\left(1-\left(\frac{3}{4}\right)^{2010}\right)khong\:làsốnguyên\)

\(A+\frac{3}{4}A=1+\left(\frac{3}{4}\right)^{2011}\)

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

\(\Leftrightarrow A=\left(1+\left(\frac{3}{4}\right)^{2011}\right):\frac{7}{4}=\frac{4}{7}\left(1+\left(\frac{3}{4}\right)^{2011}\right)\)

Vì \(1<1+\left(\frac{3}{4}\right)^{2011}<1+\frac{3}{4}=\frac{7}{4}\)

=> 4/7 < A < 4/7 .7/4 =1  =>  A không là số nguyên