\(A=1+\frac{1}{3^2}+\frac{1}{3^4}+...+\frac{1}{3^{100}}\)

\(9A=9+1+\frac{1}{3^2}+...+\frac{1}{3^{98}}\)

\(\Rightarrow8A=9-\frac{1}{3^{100}}\)

Có \(8A=9-\frac{1}{3^n}=9-\frac{1}{3^{98}}\)

\(\Rightarrow n=98\)

Quên . Phải đóng ngoặc mấy chỗ x-2

\(\frac{x-2}{12}+\frac{x-2}{20}+\frac{x-2}{30}+\frac{x+2}{42} +\frac{x+2}{56}+\frac{x+2}{72}=\frac{16}{9}\)

\(\Rightarrow x-2\left(\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+\frac{1}{7.8}+\frac{1}{8.9}\right)=\frac{16}{9}\)

\(\Rightarrow x-2\left(\frac{1}{3}+\frac{1}{4}-\frac{1}{4}+\frac{1}{5}-...+\frac{1}{8}-\frac{1}{8}+\frac{1}{9}\right)=\frac{16}{9}\)

\(\Rightarrow x-2\left(\frac{1}{3}+\frac{1}{9}\right)=\frac{16}{9}\)

\(\Rightarrow\left(x-2\right)\frac{4}{9}=\frac{16}{9}\)

\(\Rightarrow x-2=4\)

\(\Rightarrow x=6\)

Mình không chắc đâu

/5x-4/=/x+2/

TH1:5x-4=x+2

4+2=5x-x

6=4x

=>x=1,5

TH2:5x-4=-x+2

4+2=5x+x

6=6x

=>x=1

TH3:-5x-4=x+2

4+2=-5x-x

6=-6x

=>x=-1

Vậy x\(\in\)\(\left\{-1;1;1,5\right\}\)