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\(\left(1-\dfrac{1}{1+2}\right)\cdot\left(1-\dfrac{1}{1+2+3}\right)\cdot\left(\dfrac{1}{1+2+3+...+2006}\right)\)
\(=\left(1-\dfrac{1}{3}\right)\cdot\left(1-\dfrac{1}{6}\right)\cdot\left\{\dfrac{1}{\left(2006+1\right)\left[\left(2006-1\right):1+1\right]}\right\}\)
\(=\dfrac{2}{3}\cdot\dfrac{5}{6}\cdot\dfrac{1}{2007\cdot2006}\)
\(=\dfrac{10}{18}\cdot\dfrac{1}{4026042}\)
\(=\dfrac{5}{9}\cdot\dfrac{1}{4026042}\)
\(=\dfrac{5}{36234378}\)
\(N=1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\Rightarrow2N=2+1+\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+...+\left(\dfrac{1}{2}\right)^{99}\)
\(\Rightarrow N=2N-N=2+1+\dfrac{1}{2}+...+\left(\dfrac{1}{2}\right)^{99}-1-\dfrac{1}{2}-...-\left(\dfrac{1}{2}\right)^{100}=2-\left(\dfrac{1}{2}\right)^{100}\)
\(N=1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\)
\(\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\)
\(\dfrac{1}{2}N-N=\left(\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+\left(\dfrac{1}{2}\right)^4+...+\left(\dfrac{1}{2}\right)^{101}\right)\)
\(-\left(1+\left(\dfrac{1}{2}\right)+\left(\dfrac{1}{2}\right)^2+\left(\dfrac{1}{2}\right)^3+...+\left(\dfrac{1}{2}\right)^{100}\right)\)
\(-\dfrac{1}{2}N=\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}-1\)
\(N=\dfrac{-\dfrac{1}{2}+\left(\dfrac{1}{2}\right)^{101}}{-\dfrac{1}{2}}\)
\(P=\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2+\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{3}\right)^4+...+\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\)
\(=\left(\dfrac{1}{3}-\left(\dfrac{1}{3}\right)^2\right)+\left(\left(\dfrac{1}{3}\right)^3-\left(\dfrac{1}{4}\right)^4\right)+...+\left(\left(\dfrac{1}{3}\right)^{19}-\left(\dfrac{1}{3}\right)^{20}\right)\)
\(=\dfrac{1}{3}.\dfrac{2}{3}+\left(\dfrac{1}{3}\right)^3.\dfrac{2}{3}+...+\left(\dfrac{1}{3}\right)^{19}.\dfrac{2}{3}\)
\(=\dfrac{2}{3}.\left[\dfrac{1}{3}+\left(\dfrac{1}{3}\right)^3+...+\left(\dfrac{1}{3}\right)^{19}\right]\)
Bài 2:
a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)
b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)
\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)
\(=x^4-22x^3+108x^2-45x\)
c: \(=12x^5-18x^4+30x^3-24x^2\)
d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)
A = \(\left(-2\right).\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)...\left(-1\dfrac{1}{214}\right)\)
= \(\left(-2\right).\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{215}{214}\right)\)
= \(\dfrac{\left(-2\right).\left(-3\right).\left(-4\right).\left(-5\right)...\left(-215\right)}{1.2.3.4...214}\)
= \(\dfrac{2.3.4.5...215}{1.2.3.4...214}\)
= \(\dfrac{215}{1}=215\)
B = \(\left(-1\dfrac{1}{2}\right).\left(-1\dfrac{1}{3}\right).\left(-1\dfrac{1}{4}\right)....\left(-1\dfrac{1}{299}\right)\)
= \(\left(-\dfrac{3}{2}\right).\left(-\dfrac{4}{3}\right).\left(-\dfrac{5}{4}\right)...\left(-\dfrac{300}{299}\right)\)
= \(\dfrac{\left(-3\right).\left(-4\right).\left(-5\right)...\left(-300\right)}{2.3.4...299}\)
= \(\dfrac{3.4.5...300}{2.3.4.5...299}\)
= \(\dfrac{300}{2}=150\)
B=\(\dfrac{1}{2}:\left(-1\dfrac{1}{2}\right):1\dfrac{1}{3}:....:\left(-1\dfrac{1}{100}\right)\)
=\(\dfrac{1}{2}:\dfrac{-3}{2}:\dfrac{4}{3}:....:\dfrac{-101}{100}\)
=\(\dfrac{1}{2}.\dfrac{-2}{3}.\dfrac{3}{4}........\dfrac{-100}{101}\)
=\(\dfrac{1.\left(-2\right).3......\left(-100\right)}{2.3.4...........101}\)
=\(\dfrac{1}{101}\)
\(\dfrac{\left(\dfrac{-1}{2}\right)^3-\left(\dfrac{3}{4}\right)^3.\left(-2\right)^2}{2.\left(-1\right)^5+\left(\dfrac{3}{4}\right)^2-\dfrac{3}{8}}\)
\(=\dfrac{\dfrac{-1}{8}-\dfrac{27}{64}.4}{-2+\dfrac{9}{16}-\dfrac{3}{8}}=\dfrac{-\dfrac{1}{8}-\dfrac{27}{16}}{-\dfrac{29}{16}}\)
\(=\dfrac{-\dfrac{29}{16}}{-\dfrac{29}{16}}=1\)
Chúc bạn học tốt!!!
A=\(\dfrac{1\cdot4}{2\cdot3}\) \(\cdot\dfrac{2\cdot5}{3\cdot4}\) ...\(\dfrac{2015\cdot2018}{2016\cdot2017}\)
A=\(\dfrac{1\cdot2\cdot3\cdot...\cdot2015}{1\cdot2\cdot3\cdot...\cdot2016}\) \(\cdot\dfrac{4\cdot5\cdot...\cdot2018}{3.4\cdot...\cdot2017}\)
A=\(\dfrac{1}{2016}\) \(\cdot\dfrac{2018}{3}\) =\(\dfrac{1009}{336}\)