\(\left(\frac{13}{2012}-2013\right).\frac{1}{2013}-\left(\frac{1}{2012}-2013\right).\frac{13}{2013}=?\)
Những câu hỏi liên quan
So sánh \(\left(\frac{2012^{2012}}{2013^{2012}}+1\right)^{2013}\) và \(\left(\frac{2012^{2013}}{2013^{2013}}+1\right)^{2012}\)
Ta có \(\frac{2012^{2013}}{2013^{2013}}=\frac{2012^{2012}}{2013^{2012}}.\frac{2012}{2013}\)
Vì \(\frac{2012}{2013}< 1\)nên\(\frac{2012^{2012}}{2013^{2012}}.\frac{2012}{2013}< \frac{2012^{2012}}{2013^{2012}}.1=\frac{2012^{2012}}{2013^{2012}}\)
hay \(\frac{2012^{2013}}{2013^{2013}}< \frac{2012^{2012}}{2013^{2012}}\)
\(\Rightarrow\frac{2012^{2013}}{2013^{2013}}+1< \frac{2012^{2012}}{2013^{2012}}+1\)
\(\Rightarrow\left(\frac{2012^{2013}}{2013^{2013}}+1\right)^{2012}< \left(\frac{2012^{2012}}{2013^{2012}}+1\right)^{2013}\)
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giải pt
\(0,05\left(\frac{2x-2}{2011}+\frac{2x}{2012}+\frac{2x+2}{2013}\right)=3,3-\left(\frac{x-1}{2011}+\frac{x}{2012}+\frac{x+1}{2013}\right)\)
\(\left(\frac{5}{4}-\frac{2}{5}\right).\frac{2012}{2013}+\left(\frac{3}{4}-\frac{3}{5}\right).\frac{2012}{2013}\)
\(=\left(\frac{5}{4}-\frac{2}{5}+\frac{3}{4}-\frac{3}{5}\right).\frac{2012}{2013}\)
\(=\left(\frac{8}{4}-\frac{5}{5}\right).\frac{2012}{2013}\)
\(=\left(2-1\right).\frac{2012}{2013}\)
\(=\frac{2012}{2013}\)
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\(=\frac{2012}{2013}.\left(\frac{5}{4}-\frac{2}{5}+\frac{3}{4}-\frac{3}{5}\right)=\frac{2012}{2013}.\left(\frac{5}{4}+\frac{3}{4}-\frac{2}{5}-\frac{3}{5}\right)=\frac{2012}{2013}.\left(\frac{8}{4}-\frac{5}{5}\right)=\frac{2012}{2013}.\left(1-1\right)=\frac{2012}{2013}.0\)
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cho so A=\(\frac{2013+\frac{1}{2}}{\left(2012+\frac{1}{2}\right)^2+2013+\frac{1}{2}}\)
B=\(\frac{2013+\frac{1}{3}}{\left(2012+\frac{1}{3}\right)^2+2013+\frac{1}{3}}\)
so sanh A va B
Tìm x biết : \(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2004}\right)x=\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\left(\frac{2012}{2}+1\right)+...+\left(\frac{2}{2012}+1\right)+\left(\frac{1}{2013}+1\right)+1\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=\frac{2014}{2}+...+\frac{2014}{2012}+\frac{2014}{2013}+\frac{2014}{2014}\)
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right)x=2014.\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)\)
\(x=\frac{2014.\left(\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}+\frac{1}{2014}\right)}{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}}\)
\(x=2014\)
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(\(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
tìm x
Ta có: \(\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=2013+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=1+\left(1+\frac{2012}{2}\right)+...+\left(1+\frac{2}{2012}\right)+\left(1+\frac{1}{2013}\right)\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=\frac{2014}{2014}+\frac{2014}{2}+\frac{2014}{3}+...+\frac{2014}{2012}+\frac{2014}{2013}\)
\(\Rightarrow\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2014}\right)x=2014.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}+\frac{1}{2014}\right)\)
\(\Rightarrow x=2014\)
Lưu ý: số 2013 ở dòng T2 được tách ra làm 2013 số 1
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1) cho A=\(\frac{\left(x+2012\right)^2+2\left(x+2013\right)\left(x-2013\right)+\left(x-2012\right)^2}{\left(x^2-2012\right)+\left(x^2-2013\right)}\)
Tính giá trị A tại x=20162017
\(\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2014}\right).x=\frac{2013}{1}+\frac{2012}{2}+...+\frac{2}{2012}+\frac{1}{2013}\)Tìm x biết:
Tìm số nguyên \(x\)nhỏ nhất thỏa mãn:
\(\left(\frac{2011}{2012}+\frac{2012}{2013}+\frac{2013}{2011}\right).\left(x-2013\right)>3x-6039\)