Đặt A \(=\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\)

\(\Rightarrow3A=1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\)

\(\Rightarrow3A-A=\left(1+\dfrac{2}{3}+\dfrac{3}{3^2}+...+\dfrac{100}{3^{99}}\right)-\left(\dfrac{1}{3}+\dfrac{2}{3^2}+\dfrac{3}{3^3}+...+\dfrac{100}{3^{100}}\right)\)

\(\Rightarrow2A=1+\left(\dfrac{1}{3}+\dfrac{1}{3^2}+...+\dfrac{1}{3^{99}}\right)-\dfrac{100}{3^{100}}\)

\(\Rightarrow2A=1+\dfrac{1}{2}-\dfrac{1}{3^{99}\times2}-\dfrac{100}{3^{100}}\)

\(\Rightarrow A=\dfrac{3}{4}-\dfrac{1}{3^{99}\times4}-\dfrac{50}{3^{100}}< \dfrac{3}{4}\)

Vậy ...

Chúc Các Bạn Học Tốt !

a, \(\dfrac{5}{3}+\left(\dfrac{-2}{7}\right)-\left(\dfrac{-1}{2}\right)\)

\(=\dfrac{5}{3}-\dfrac{2}{7}+\dfrac{1}{2}\)

\(=\dfrac{29}{21}+\dfrac{1}{2}\)

\(=\dfrac{79}{42}\)

b, \(\dfrac{-4}{9}+\left(\dfrac{-5}{6}\right)-\dfrac{17}{4}\)

=\(\dfrac{-4}{9}-\dfrac{5}{6}-\dfrac{17}{4}\)

\(=\dfrac{-23}{8}-\dfrac{17}{4}\)

\(\dfrac{-199}{36}\)

ab=19

nhớ tick cho mình nha

\(3^x\times2^x=216\)

\(\Rightarrow\left(3\times2\right)^x=216\)

\(\Rightarrow6^x=216\)

\(\Rightarrow6^x=6^3\)

\(\Rightarrow x=3\)

Vậy \(x=3\)

Ta có : \(2\times1+2\times2+2\times3+...+2\times x=156\)

\(=2\times\left(1+2+3+...+x\right)=156\)

\(=1+2+3+...+x+\dfrac{156}{2}\)

\(=1+2+3+...+x=78\)

\(=\dfrac{\left(x-1\right)+1}{2}\times\left(x+1\right)=78\)

\(=\dfrac{x}{2}\times\left(x+1\right)=78\)

\(=x\times x+1=156\)

\(x\times x+1=12\times13\)

Vậy \(x=12\)

\(\left(\dfrac{7}{2}\times\dfrac{-4}{5}+\dfrac{7}{2}\times\dfrac{-1}{5}\right)\times20\%-0,1\)

\(=\dfrac{7}{2}\left(\dfrac{-4}{5}+\dfrac{-1}{5}\right)\times20\%-0,1\)

\(=\dfrac{7}{2}\times-1\times20\%-0,1\)

\(=\dfrac{-7}{2}\times20\%-0,1\)

\(=-0,7-0,1\)

\(=-0,8\)

a. \(A=\left\{4,5,6\right\}\)

b. \(B=\left\{4,6,9,11\right\}\)

Sửa lại chỗ :" Gấn tận cơ nhị đầu " thành " Gân tân cơ nhị đầu "