`1//([-1]/2)^2 . |+8|-(-1/2)^3:|-1/16|=1/4 .8+1/8 .16=2+2=4`

`2//|-0,25|-(-3/2)^2:1/4+3/4 .2017^0=0,25-2,25.4+0,75.1=0,25-9+0,75=-8,75+0,75-8`

`3//|2/3-5/6|.(3,6:2 2/5)^3=|-1/6|.(3/2)^3=1/6 . 27/8=9/16`

`4//|(-0,5)^2+7/2|.10-(29/30-7/15):(-2017/2018)^0=|1/4+7/2|.10-1/2:1=|15/4|.10-1/2=15/4 .10-1/2=75/2-1/2=37`

`5// 8/3+(3-1/2)^2-|[-7]/3|=8/3+(5/2)^2-7/3=8/3+25/4-7/3=107/12-7/3=79/12`

\(A=\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2015}-1\right)\left(\dfrac{1}{2016}-1\right)\left(\dfrac{1}{2017}-1\right)\\ A=\left(-\dfrac{1}{2}\right).\left(-\dfrac{2}{3}\right).\left(-\dfrac{3}{4}\right)...\left(-\dfrac{2014}{2015}\right)\left(-\dfrac{2015}{2016}\right)\left(-\dfrac{2016}{2017}\right)\\ A=\dfrac{1.2.3.4...2014.2015.2016}{2.3.4...2015.2016.2017}=\dfrac{1}{2017}\)

\(B=\left(-1\dfrac{1}{2}\right)\left(-1\dfrac{1}{3}\right)\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{2015}\right)\left(-1\dfrac{1}{2016}\right)\left(-1\dfrac{1}{2017}\right)\\ B=\left(-\dfrac{3}{2}\right)\left(-\dfrac{4}{3}\right)\left(-\dfrac{5}{4}\right)...\left(-\dfrac{2016}{2015}\right)\left(-\dfrac{2017}{2016}\right)\left(-\dfrac{2018}{2017}\right)\\ B=\dfrac{3.4.5...2016.2017.2018}{2.3.4...2015.2016.2017}=\dfrac{2018}{2}=1009\)

\(M=A.B=\dfrac{1}{2017}.1009=\dfrac{1009}{2017}\)

\(A=\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-2016}{2017}=\dfrac{1}{2017}\)

\(B=\dfrac{-3}{2}\cdot\dfrac{-4}{3}\cdot...\cdot\dfrac{-2018}{2017}=\dfrac{2018}{2}=1009\)

\(A\cdot B=\dfrac{1009}{2017}\)

c)

Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)

\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)

d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)

\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)

\(=3-1+\dfrac{1}{4}:2\)

\(=3-1+\dfrac{1}{8}\)

\(=\dfrac{17}{8}\)

f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)

\(=\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{5}{8}\)

\(=\dfrac{3}{8}+\dfrac{5}{8}\)

\(=1\)

quy đồng lên ta đc \(\dfrac{2^2-1}{2^2}......\dfrac{2017^2-1}{2017^2}\)

khai triển hằng đẳng thức \(a^2-b^2=\left(a-b\right)\left(a+b\right)\)ta đc\(\dfrac{1.3}{2^2}.\dfrac{2.4}{3^2}.....\dfrac{2016.2018}{2017}\)

=\(\dfrac{1.2.3^2.4^2.....2016^2.2017.2018}{2^2.......2017^2}=\dfrac{1.2018}{2.2017}=\dfrac{1009}{2017}\)

Bài 2 :

\(S=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+............+\dfrac{2017}{4^{2017}}\)

\(\Leftrightarrow4S=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...........+\dfrac{2017}{4^{2016}}\)

\(\Leftrightarrow4S-S=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+..........+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+..........+\dfrac{2017}{4^{2017}}\right)\)

\(\Leftrightarrow3S=1+\dfrac{1}{4}+\dfrac{1}{4^2}+.........+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2016}}\)

Đặt :

\(A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2016}}\)

\(\Leftrightarrow4A=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+..........+\dfrac{1}{4^{2015}}\)

\(\Leftrightarrow4A-A=\left(4+1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+.......+\dfrac{1}{4^{2016}}\right)\)

\(\Leftrightarrow3A=4-\dfrac{1}{4^{2016}}\)

\(\Leftrightarrow D=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}\)

\(\Leftrightarrow3S=\dfrac{4}{3}-\dfrac{1}{2^{2016}.3}-\dfrac{2017}{4^{2016}}\)

\(\Leftrightarrow3S< \dfrac{4}{3}\)

\(\Leftrightarrow S< \dfrac{4}{9}\)

\(\Leftrightarrow S< \dfrac{1}{2}\rightarrowđpcm\)

\(A=\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\) ( A cho đẹp :v)

\(4A=4\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)

\(4A=1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\)

\(4A-A=\left(1+\dfrac{2}{4}+\dfrac{3}{4^2}+...+\dfrac{2017}{4^{2016}}\right)-\left(\dfrac{1}{4}+\dfrac{2}{4^2}+\dfrac{3}{4^3}+...+\dfrac{2017}{4^{2017}}\right)\)\(3A=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}-\dfrac{2017}{4^{2017}}\)

Đặt:

\(M=1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\)

\(4M=4\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)

\(4M=4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\)

\(4M-M=\left(4+1+\dfrac{1}{4}+\dfrac{1}{4^2}+...+\dfrac{1}{4^{2015}}\right)-\left(1+\dfrac{1}{4}+\dfrac{1}{4^2}+\dfrac{1}{4^3}+...+\dfrac{1}{4^{2016}}\right)\)\(3M=4-\dfrac{1}{4^{2016}}\)

\(M=\dfrac{4}{3}-\dfrac{1}{4^{2016}}\)

Thay M vào A ta có:

\(A=\dfrac{4}{9}-\dfrac{1}{4^{2016}.3}-\dfrac{2017}{4^{2017}}\)

\(\Rightarrow A< \dfrac{1}{2}\Rightarrowđpcm\)

\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2017}-1\right)\\ =\dfrac{-1}{2}\cdot\dfrac{-2}{3}\cdot\dfrac{-3}{4}\cdot...\cdot\dfrac{-2016}{2017}\\ =\dfrac{\left(-1\right)\cdot\left(-2\right)\cdot\left(-3\right)\cdot...\cdot\left(-2016\right)}{2\cdot3\cdot4\cdot...\cdot2017}\\=\dfrac{\left(-1\right)\cdot\left(-1\right)\cdot\left(-1\right)\cdot...\left(-1\right)}{2017}\left(\text{có 2016 thừa số -1}\right)\\ =\dfrac{1}{2017}\)