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\(a)\) Ta có :
\(VP=\frac{2018}{1}+\frac{2017}{2}+\frac{2016}{3}+...+\frac{2}{2017}+\frac{1}{2018}\)
\(VP=\left(\frac{2018}{1}-1-...-1\right)+\left(\frac{2017}{2}+1\right)+\left(\frac{2016}{3}+1\right)+...+\left(\frac{2}{2017}+1\right)+\left(\frac{1}{2018}+1\right)\)
\(VP=1+\frac{2019}{2}+\frac{2019}{3}+...+\frac{2019}{2017}+\frac{2019}{2018}\)
\(VP=2019\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2017}+\frac{1}{2018}+\frac{1}{2019}\right)\)
Lại có :
\(VT=\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2019}\right).x\)
\(\Rightarrow\)\(x=2019\)
Vậy \(x=2019\)
Chúc bạn học tốt ~
\(\frac{x+4}{2015}+\frac{x+3}{2016}=\frac{x+2}{2017}+\frac{x+1}{2018}\)
\(\Rightarrow\frac{x+4}{2015}+1+\frac{x+3}{2016}+1=\frac{x+2}{2017}+1+\frac{x+1}{2018}+1\)
\(\Rightarrow\frac{x+4+2015}{2015}+\frac{x+3+2016}{2016}=\frac{x+2+2017}{2017}+\frac{x+1+2018}{2018}\)
\(\Rightarrow\frac{x+2019}{2015}+\frac{x+2019}{2016}-\frac{x+2019}{2017}-\frac{x+2019}{2018}=0\)
\(\Rightarrow\left(x+2019\right)\left(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\right)=0\)
Vì \(\frac{1}{2015}+\frac{1}{2016}-\frac{1}{2017}-\frac{1}{2018}\ne0\)
=> x + 2019 = 0
=> x = -2019
Vậy x = -2019
Em chuyển sang cùng một vế rồi ghép cái đầu vói cái thứ 3 cái thứ 2 với cái cuối. :)Dùng quy đồng :)
\(\frac{1}{1.3}+\frac{1}{3.5}+...+\frac{1}{x.\left(x+2\right)}=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{3}+\frac{1}{3}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow\frac{1}{2}.\left(1-\frac{1}{x+2}\right)=\frac{20}{41}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{20}{41}\div\frac{1}{2}\)
\(\Leftrightarrow1-\frac{1}{x+2}=\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=1-\frac{40}{41}\)
\(\Leftrightarrow\frac{1}{x+2}=\frac{1}{41}\)
\(\Leftrightarrow x+2=41\)
\(\Leftrightarrow x=41-2\)
\(\Leftrightarrow x=39\)
P \(=\left(1-\frac{1}{2^2}\right).\left(1-\frac{1}{3^2}\right).\left(1-\frac{1}{4^2}\right)...\left(1-\frac{1}{50^2}\right)\)
P\(=\frac{2^2-1}{2^2}.\frac{3^2-1}{3^2}.\frac{4^2-1}{4^2}...\frac{50^2-1}{50^2}\)
P \(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}...\frac{49.51}{50.50}\)
P\(=\frac{\left(1.2.3...49\right).\left(3.4.5...51\right)}{\left(2.3.4...50\right).\left(2.3.4...50\right)}\)
P\(=\frac{1.51}{50.2}=\frac{51}{100}\)
ta có:
\(\frac{x-2016}{2015}+\frac{x-2017}{2016}-\frac{2018-x}{2017}=-3\)
\(\Leftrightarrow\left(\frac{x-2016}{2015}+1\right)+\left(\frac{x-2017}{2016}+1\right)+\left(\frac{x-2018}{2017}+1\right)=0\)
\(\Leftrightarrow\frac{x-1}{2015}+\frac{x-1}{2016}+\frac{x-1}{2017}=0\)
\(\Leftrightarrow\left(x-1\right)\left(\frac{1}{2015}+\frac{1}{2016}+\frac{1}{2017}\right)=0\)
\(\Rightarrow x-1=0\Leftrightarrow x=1\)
\(\frac{x+4}{2016}+\frac{x+3}{2017}=\frac{x+2}{2018}+\frac{x+1}{2019}\)
\(\Rightarrow\frac{x+4}{2016}+1+\frac{x+3}{2017}+1=\frac{x+2}{2018}+1+\frac{x+1}{2019}+1\)
\(\Rightarrow\frac{x+4+2016}{2016}+\frac{x+3+2017}{2017}=\frac{x+2+2018}{2018}+\frac{x+1+2019}{2019}\)
\(\Rightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}=\frac{x+2020}{2018}+\frac{x+2020}{2019}\)
\(\Rightarrow\frac{x+2020}{2016}+\frac{x+2020}{2017}-\frac{x+2020}{2018}-\frac{x+2020}{2019}=0\)
\(\Rightarrow\left(x+2020\right)\left(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}\right)=0\)
\(\Rightarrow x+2020=0\) vì \(\frac{1}{2016}+\frac{1}{2017}-\frac{1}{2018}-\frac{1}{2019}>0\)
\(\Rightarrow x=-2020\)
uk