\(A=1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2011}}+\dfrac{1}{2^{2012}}\)

\(\Rightarrow2A=2+1+\dfrac{1}{2}+...+\dfrac{1}{2^{2011}}\)

\(\Rightarrow2A-A=2-\dfrac{1}{2^{2012}}\)

\(\Rightarrow A=2-\dfrac{1}{2^{2012}}\)

\(A= 1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\)\(\dfrac{1}{2^{2012}}\)

⇒\(2A=2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\)

⇒\(2A-A=(2+1+\dfrac{1}{2}+...+\)\(\dfrac{1}{2^{2012}}\))\(-\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+...+\dfrac{1}{2^{2012}}\right)\)

⇒\(A=2-\)\(\dfrac{1}{2^{2012}}\)

\(A=3\left|1-2x\right|-5\)

Ta có: \(\left|1-2x\right|\ge0\forall x\)

\(\Rightarrow3.\left|1-2x\right|-5\ge-5\forall x\)

\(\Rightarrow A\ge-5\forall x\)

Dấu "=" xảy ra

\(\Leftrightarrow3.\left|1-2x\right|=0\Leftrightarrow1-2x=0\Leftrightarrow x=\dfrac{1}{2}\)

\(a,=\dfrac{5}{6}+\dfrac{6}{7}\\ =\dfrac{35}{42}+\dfrac{36}{42}=\dfrac{71}{42}\\ b,=\dfrac{5}{6}+\dfrac{4}{9}\\ =\dfrac{15}{18}+\dfrac{8}{18}=\dfrac{23}{18}\)

\(A=2x+xy^2-x^2y-2y\)

\(=2\left(x-y\right)-xy\left(x-y\right)\)

\(=\left(x-y\right)\left(2-xy\right)\)

\(=\left(-\dfrac{1}{2}-\dfrac{-1}{3}\right)\left(2-\dfrac{-1}{2}\cdot\dfrac{-1}{3}\right)\)

\(=\left(\dfrac{1}{3}-\dfrac{1}{2}\right)\cdot\left(2-\dfrac{1}{6}\right)\)

\(=\dfrac{-1}{6}\cdot\dfrac{11}{6}=-\dfrac{11}{36}\)

\(tanx=\dfrac{1}{2}\Leftrightarrow\dfrac{sinx}{cosx}=\dfrac{1}{2}\Leftrightarrow cosx=2sinx\)

\(1+tan^2x=\dfrac{1}{cos^2x}\) \(\Leftrightarrow cos^2x=\dfrac{4}{5}\)

=> \(sin2x=2sinx.cosx=cos^2x\)

\(A=\dfrac{2sin2x}{2-3cos2x}=\dfrac{2cos^2x}{2-3\left(cos^2x-1\right)}=\dfrac{8}{13}\)

a: \(2\dfrac{3}{5}+1\dfrac{2}{5}\cdot\dfrac{31}{2}\)

\(=\dfrac{13}{5}+\dfrac{7}{5}\cdot\dfrac{31}{2}\)

\(=\dfrac{26}{10}+\dfrac{217}{10}=\dfrac{243}{10}\)

b: \(4\dfrac{3}{4}-3\dfrac{2}{3}:1\dfrac{1}{6}\)

\(=\dfrac{19}{4}-\dfrac{11}{3}:\dfrac{7}{6}\)

\(=\dfrac{19}{4}-\dfrac{11}{3}\cdot\dfrac{6}{7}\)

\(=\dfrac{19}{4}-\dfrac{22}{7}\)

\(=\dfrac{19\cdot7-22\cdot4}{28}=\dfrac{45}{28}\)