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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 \(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\ne0\)
\(\Rightarrow x=-2020\)
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
Đặt \(S=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}{\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}}\)
Biến đổi mẫu
\(\frac{2017}{1}+\frac{2016}{2}+...+\frac{1}{2017}\)
\(=\left(2017+1\right)+\left(\frac{2016}{2}+1\right)+...+\left(\frac{1}{2017}+1\right)-2017\)
\(=2018+\frac{2018}{2}+...+\frac{2018}{2017}+\frac{2018}{2018}-2018\)
\(=2018.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)\)
\(\Rightarrow S=\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}}{2018.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2018}\right)}=\frac{1}{2018}\)
\(\frac{x-1}{2018}+\frac{x-2}{2017}=\frac{x-3}{2016}+\frac{x-4}{2015}\)
\(\Rightarrow\frac{x-1}{2018}-1+\frac{x-2}{2017}-1=\frac{x-3}{2016}-1+\frac{x-4}{2015}-1\)
\(\Rightarrow\frac{x-1-2018}{2018}+\frac{x-2-2017}{2017}=\frac{x-3-2016}{2016}+\frac{x-4-2015}{2015}\)
\(\Rightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}=\frac{x-2019}{2016}+\frac{x-2019}{2015}\)
\(\Rightarrow\frac{x-2019}{2018}+\frac{x-2019}{2017}-\frac{x-2019}{2016}-\frac{x-2019}{2015}=0\)
\(\Rightarrow\left(x-2019\right)\left(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\right)=0\)
Mà \(\frac{1}{2018}+\frac{1}{2017}-\frac{1}{2016}-\frac{1}{2015}\ne0\)
\(\Rightarrow x-2019=0\)
\(\Rightarrow x=2019\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\frac{12\left(x+2015\right)}{60}+\frac{15\left(x+2016\right)}{60}=\frac{20\left(x+2017\right)}{60}+\frac{30\left(x+2018\right)}{60}\)
\(\Rightarrow12x+24180+15x+30240=20x+40340+30x+60540\)
\(\Leftrightarrow-23x=22460\Leftrightarrow x=-\frac{22460}{23}\)
Đặt \(A=4^{2018}-4^{2017}+4^{2016}-4^{2015}+.......+4^2-4\)
\(\Rightarrow4A=4^{2019}-4^{2018}+4^{2017}-4^{2016}+......+4^3-4^2\)
\(\Rightarrow4A+A=5A=4^{2019}-4\)\(\Rightarrow A=\frac{4^{2019}-4}{5}\)