Tính M \(\sqrt{1+1013^2+\frac{1013^2}{1014^2}}+\frac{1013}{1014}\)
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Mk sửa 1013 thành 1008 nhá
\(\frac{x-2}{2015}+\frac{x-3}{2014}=\frac{x-1}{1008}\)
\(\Leftrightarrow\frac{x-2}{2015}+\frac{x-3}{2014}-2=\frac{x-1}{1008}-2\)
\(\Leftrightarrow\left(\frac{x-2}{2015}-1\right)+\left(\frac{x-3}{2014}-1\right)=\frac{x-1}{1013}-2\)
\(\Leftrightarrow\frac{x-2-2015}{2015}+\frac{x-3-2014}{2014}=\frac{x-1-2016}{1008}\)
\(\Leftrightarrow\frac{x-2017}{2015}+\frac{x-2017}{2014}=\frac{x-2017}{1008}\)
\(\Leftrightarrow\frac{x-2017}{2015}+\frac{x-2017}{2014}-\frac{x-2017}{1008}=0\)
\(\Leftrightarrow\left(x-2017\right)\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}\right)=0\)
\(\Leftrightarrow x-2017=0\times\left(\frac{1}{2015}+\frac{1}{2014}-\frac{1}{1008}\right)\)
\(\Leftrightarrow x-2017=0\)
\(\Leftrightarrow x=2017\)
Hok TOT ^_^
Ta có : Q=\(\frac{1010+1011+1012}{1011+1012+1013}\)=\(\frac{1010}{1011+1012+1013}+\frac{1011}{1011+1012+1013}+\frac{1012}{1011+1012+1013}\)
Vì1010/1011>1010/1011+1012+1013
1011/1012>1011/1011+1012+1013
1012/1013>1012/1011+1012+1013
=>P>Q
\(\frac{x-1}{2016}+\frac{x-2}{2015}-\frac{x-3}{2014}=\frac{x-4}{2013}\)
\(\left(\frac{x-1}{2016}-1\right)+\left(\frac{x-2}{2016}-1\right)-\left(\frac{x-3}{2014}-1\right)=\left(\frac{x-4}{2013}-1\right)\)
\(\frac{x-2017}{2016}+\frac{x-2017}{2015}-\frac{x-2017}{2014}=\frac{x-2017}{2013}\)
\(\frac{x-2017}{2016}+\frac{x-2017}{2015}+\frac{x-2017}{2014}-\frac{x-2017}{2013}=0\)
\(\left(x-2017\right)\left(\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\right)=0\)
\(x-2017=0\left(vì\frac{1}{2016}+\frac{1}{2015}-\frac{1}{2014}-\frac{1}{2013}\ne0\right)\)
x=2017
\(\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-\frac{1014}{1014}\right).\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...\left(1-1\right).\left(1-\frac{1015}{1014}\right)\)
\(=\left(1-\frac{1}{1014}\right).\left(1-\frac{2}{1014}\right).\left(1-\frac{3}{1014}\right).\left(1-\frac{4}{1014}\right)...0.\left(1-\frac{1015}{1014}\right)\)
\(=0\)
Đặt a=2013
\(\Rightarrow M=\sqrt{1+a^2+\frac{a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(\Rightarrow M=\sqrt{\frac{\left(a+1\right)^2+a^2\left(a+1\right)^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(\Rightarrow M=\sqrt{\frac{a^2+2a+1+a^4+2a^3+a^2+a^2}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(\Rightarrow M=\sqrt{\frac{\left(a^4+2a^3+a^2\right)+2\left(a^2+a\right)+1}{\left(a+1\right)^2}}+\frac{a}{a+1}\)
\(\Rightarrow M=\sqrt{\left(\frac{a^2+a+1}{a+1}\right)^2}+\frac{a}{a+1}\)
\(\Rightarrow M=\frac{a^2+a+1+a}{a+1}\)(Bỏ trị tuyệt đối vì a=2013)
\(\Rightarrow M=\frac{a^2+2a+1}{a+1}=\frac{\left(a+1\right)^2}{a+1}=a+1=1013+1=1014\)