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13 tháng 8 2019
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 !!
13 tháng 8 2019
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
TN
5 tháng 7 2016
x+2015/5 + x+2016/4=x+2017/3 + x+2018/2
\(\Rightarrow\frac{x+2015}{5}+1+\frac{x+2016}{4}+1=\frac{x+2017}{3}+1+\frac{x+2018}{2}+1\)
\(\Rightarrow\frac{x+2020}{5}+\frac{x+2020}{4}=\frac{x+2020}{3}+\frac{x+2020}{2}\)
\(\Rightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Rightarrow x+2020=0\).Do \(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)
\(\Rightarrow x=-2020\)
10 tháng 7 2017
a) \(\frac{x+2015}{5}+\frac{x+2015}{6}=\frac{x+2015}{7}+\frac{x+2015}{8}\)
\(\frac{x+2015}{5}+\frac{x+2015}{6}-\frac{x+2015}{7}-\frac{x+2015}{8}=0\)
\(\left(x+2015\right).\left(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\right)=0\)
vì \(\frac{1}{5}+\frac{1}{6}-\frac{1}{7}-\frac{1}{8}\ne0\)
\(\Rightarrow\)x + 2015 = 0
\(\Rightarrow\)x = -2015
b) Tương tự
LV
0
5 tháng 5 2017
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
TC
1
4 tháng 12 2018
\(\frac{x-1}{2018}+\frac{x-2}{2017}=\frac{x-3}{2016}+\frac{x-4}{2015}\)
\(\Rightarrow\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)
\(\Rightarrow\frac{x-1-2018}{2018}+\frac{x-2-2017}{2017}=\frac{x-3-2016}{2016}+\frac{x-4-2015}{2015}\)
\(\Rightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}=\frac{x-2019}{2016}+\frac{x-2019}{2015}\)
\(\Rightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}-\frac{x-2019}{2016}-\frac{x-2019}{2015}=0\)
\(\Rightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
Mà \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\)
\(\Rightarrow x-2019=0\)
\(\Rightarrow x=2019\)