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x=2016 =>x-1=2015
Suy ra: \(C=x^{2010}-2015x^{2009}-2015x^{2008}-...-2015x+1\)
\(=x^{2010}-\left(x-1\right).x^{2009}-\left(x-1\right).x^{2008}-...-\left(x-1\right).x+1\)
\(=x^{2010}-x^{2010}+x^{2009}-x^{2009}+x^{2008}-...-x^2+x+1\)
\(=x+1=2016+1=2017\)
\(\frac{\frac{2017}{1}+\frac{2016}{2}+\frac{2015}{3}+...+\frac{1}{2017}+2018}{\frac{1}{1}+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(=\frac{1+\left(\frac{2016}{2}+1\right)+\left(\frac{2015}{3}+1\right)+...+\left(\frac{1}{2017}+1\right)+2018}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(=\frac{\frac{2018}{2018}+\frac{2018}{2}+\frac{2018}{3}+...+\frac{2018}{2017}+\frac{2018}{1}}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
\(=\frac{2018.\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}\right)}{1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}\)
= 2018
Ta có: \(N\left(x\right)=x^{2017}-2018x^{2016}+2018x^{2015}-...-2018x^2+2018x-1\)
\(=x^{2017}-2018\left(x^{2016}-x^{2015}+...+x^2-x\right)-1\)
\(\Rightarrow N\left(2017\right)=2017^{2017}-2018\left(2017^{2016}-2017^{2015}+...+2017^2-2017\right)-1\)
Đặt \(A=2017^{2016}-2017^{2015}+...+2017^2-2017\)
\(\Rightarrow2017A=2017^{2017}-2017^{2016}+...+2017^3-2017^2\)
\(\Rightarrow2018A=2017^{2017}-2017\)
\(\Rightarrow A=\dfrac{2017^{2017}-2017}{2018}\)
\(\Rightarrow N\left(2017\right)=2017^{2017}-2018.\dfrac{2017^{2017}-2017}{2018}-1\)
\(=2017^{2017}-\left(2017^{2017}-2017\right)-1\)
\(=2017^{2017}-2017^{2017}+2017-1\)
\(=2016\)
Vậy N(2017) = 2016
Tìm \(x\varepsilonℝ\) biết \(\frac{x+1}{2018}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{3x+12}{2015}\)
\(\dfrac{x+1}{2018}+\dfrac{x+2}{2017}+\dfrac{x+3}{2016}=\dfrac{3x+12}{2015}\)
\(\Rightarrow\dfrac{x+1}{2018}+\dfrac{x+2}{2017}+\dfrac{x+3}{2016}-\dfrac{3x+12}{2015}=0\)
\(\Rightarrow\dfrac{x+1}{2018}+\dfrac{x+2}{2017}+\dfrac{x+3}{2016}+\dfrac{3\cdot\left(x+4\right)}{2015}=0\)
\(\Rightarrow\left(\dfrac{x+1}{2018}+1\right)+\left(\dfrac{x+2}{2017}+1\right)+\left(\dfrac{x+3}{2016}+1\right)+\left(\dfrac{3}{2015}\cdot\left(\dfrac{x+4}{2015}+1\right)\right)=0\)
\(\Rightarrow\left(x+2019\right)\cdot\left(\dfrac{1}{2018}+\dfrac{1}{2017}+\dfrac{1}{2016}+\left(\dfrac{3}{2015}\cdot\dfrac{1}{2005}\right)\right)=0\)
\(\Rightarrow x+2019=0\\ \Rightarrow x=-2019\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
Tìm \(x\varepsilonℝ\) biết \(\frac{x+1}{2018}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{3x+12}{2015}\)
\(\frac{x+1}{2018}+1+\frac{x+2}{2017}+1+\frac{x+3}{2016}+1=\frac{3x+12}{2015}+3\)
\(\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}=\frac{3 \left(x+2019\right)}{2015}\)
\(\left(x+2019\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{3}{2015}\right)=0\)
mà \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{3}{2015}\ne0\Rightarrow x+2019=0\Leftrightarrow x=-2019\)
a/ Với \(x=2016\Rightarrow2017=x+1\)
\(A=x^6-\left(x+1\right)x^5+\left(x+1\right)x^4-\left(x+1\right)x^3+\left(x+1\right)x^2-\left(x+1\right)x+2025\)
\(A=x^6-x^6-x^5+x^5+x^4-x^4-x^3+x^3+x^2-x^2-x+2025\)
\(A=2025-x=9\)
b/ Với \(x=-1\Rightarrow\left\{{}\begin{matrix}x^{2k}=1\\x^{2k+1}=-1\end{matrix}\right.\) ta có:
\(Q=2017-2016+2015-2014+...+3-2+1\)
\(Q=1+1+1+...+1+1\) (có \(\frac{2016}{2}+1=1009\) số 1)
\(Q=1009\)
\(x^{2018}+2x^{2017}+3x^{2016}+...+2017x+2018\)
\(=1+2+3+...+2017+2018\)
\(=\frac{2018.\left(2018+1\right)}{2}=2037171\)