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\(E\left(x\right)=x^{2018}-2019x^{2017}+2019x^{2016}-2019x^{2015}+...+2019x^2-2019x+1\)
Vì \(E\left(2018\right)\) nên :
\(\Rightarrow E\left(x\right)=2018^{2018}-2019.2018^{2017}+2019.2018^{2016}-2019.2018^{2015}+...+2019.2018^2-2019.2018+1\)
Tới đoạn này thì ghi dấu "=" rồi tính và làm tương tự
Lời giải
Ta có:
\(E(x)=x^{2018}-2019x^{2017}+2019x^{2016}-2019x^{2015}+...+2019x^2-2019x+1\)
\(E(x)=(x^{2018}-2018x^{2017})-(x^{2017}-2018x^{2016})+(x^{2016}-2018x^{2015})-....+(x^2-2018x)-x+1\)
\(E(x)=x^{2017}(x-2018)-x^{2016}(x-2018)+x^{2015}(x-8)-...+x(x-2018)-x+1\)
\(E(x)=(x-2018)(x^{2017}-x^{2016}+x^{2015}-...+x)-x+1\)
Suy ra \(E(2018)=-2018+1=-2017\)
=> (x+2020)/5=(x+2020)/6=(x+2020)/3+(x+2020)/2
=>(x+2020)(1/5+1/6)=(x+2020)(1/3+1/2)
Với x+2020=0=>x=-2020
Với x+2020 khác 0=>1/5+1/6=1/3+1/2 ,vô lí
Vậy x=-2020
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{x+2015}{5}+\frac{5}{5}+\frac{x+2016}{4}+\frac{4}{4}=\frac{x+2017}{3}+\frac{3}{3}+\frac{x+2018}{2}+\frac{2}{2}\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2002}{2}\)
\(\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right).\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)
\(\Leftrightarrow x=-2020\)
Vậy : \(x=-2020\)
Chúc bạn học tốt !!
a) \(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\\ \left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\\ \frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\\ \frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\\ \left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\\ \Rightarrow x+2020=0\\ \Rightarrow x=-2020\)
Vậy x = -2020
b) \(\frac{x+2015}{5}+\frac{x+2016}{6}=\frac{x+2017}{7}+\frac{x+2018}{8}\\ \left(\frac{x+2015}{5}-1\right)+\left(\frac{x+2016}{6}-1\right)=\left(\frac{x+2017}{7}-1\right)+\left(\frac{x+2018}{8}-1\right)\\ \frac{x+2010}{5}+\frac{x+2010}{6}=\frac{x+2010}{7}+\frac{x+2010}{8}\\ \frac{x+2010}{5}+\frac{x+2010}{6}-\frac{x+2010}{7}-\frac{x+2010}{8}=0\\ \left(x+2010\right)\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\\ \Rightarrow x+2010=0\\ \Rightarrow x=-2010\)
Vậy x = -2010
\(\frac{x+1}{2018}+\frac{x+2}{2017}+\frac{x+3}{2016}=\frac{x+4}{2015}+\frac{x+5}{2014}+\frac{x+6}{2013}\)
\(\Leftrightarrow\) \(\frac{x+1}{2018}+1+\frac{x+2}{2017}+1+\frac{x+3}{2016}+1=\frac{x+4}{2015}+1+\frac{x+5}{2014}+1+\frac{x+6}{2013}+1\)
\(\Leftrightarrow\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}=\frac{x+2019}{2015}+\frac{x+2019}{2014}+\frac{x+2019}{2013}\)
\(\Leftrightarrow\frac{x+2019}{2018}+\frac{x+2019}{2017}+\frac{x+2019}{2016}-\frac{x+2019}{2015}-\frac{x+2019}{2014}-\frac{x+2019}{2013}=0\)
\(\Leftrightarrow\left(x+2019\right)\left(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)\)\(=0\)
Lại có: \(\frac{1}{2018}+\frac{1}{2017}+\frac{1}{2016}-\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\) \(\ne\) \(0\)
\(\Rightarrow x+2019=0\)
\(\Rightarrow x=0-2019=-2019\)
Vậy x= -2019
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
tks bạn!!