tim x
[x+1]+[x+3]+...+[x+2017]=10092+6051
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Dat \(A=\frac{x^4+y^4}{x^4-y^4}-\frac{xy}{x^2-y^2}+\frac{x+y}{2\left(x-y\right)}\)
\(=\frac{2x^4+2y^4-2xy\left(x^2+y^2\right)+\left(x+y\right)^2\left(x^2+y^2\right)}{2x^4-2y^4}\)
\(=\frac{2x^4+2y^4+\left(x^2+y^2\right)\left[\left(x+y\right)^2-2xy\right]}{2x^4-2y^4}\)
\(=\frac{2x^4+2y^4+\left(x^2+y^2\right)^2}{2x^4-2y^4}\)
\(\Rightarrow A\ge\frac{2x^4+x^4}{2x^4}=\frac{3}{2}\)
\(\Rightarrow P=2017A\ge2017.\frac{3}{2}=\frac{6051}{2}\)
Dau '=' xay ra khi \(y=0\)
\(\frac{x+2015}{5}+\frac{x+2016}{4}=\frac{x+2017}{3}+\frac{x+2018}{2}\)
\(\Leftrightarrow\left(\frac{x+2015}{5}+1\right)+\left(\frac{x+2016}{4}+1\right)=\left(\frac{x+2017}{3}+1\right)+\left(\frac{x+2018}{2}+1\right)\)
\(\Leftrightarrow\frac{x+2020}{5}+\frac{x+2020}{4}-\frac{x+2020}{3}-\frac{x+2020}{2}=0\)
\(\Leftrightarrow\left(x+2020\right)\left(\frac{1}{5}+\frac{1}{4}-\frac{1}{3}-\frac{1}{2}\right)=0\)
\(\Leftrightarrow x+2020=0\)vì \(\frac{1}{5}+\frac{1}{4}+\frac{1}{3}+\frac{1}{2}\ne0\)
\(\Leftrightarrow x=-2020\)
\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2017}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2017}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2017}:2\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4034}\)
\(\frac{1}{x+1}=\frac{1}{2017}\)
=>x+1=2017
=>x=2016
\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2015}{2016}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(2\left(\frac{1}{2}-\frac{1}{x+1}\right)=\frac{2015}{2016}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2016}:2\)
\(\frac{1}{x+1}=\frac{1}{2}-\frac{2015}{4032}\)
\(\frac{1}{x+1}=\frac{1}{4032}\)
=>x+1=4032
=>x=4031
Min A, Min B chứ em.
Định làm nhưng lười với quên béng mất cách trình bày ==''