\(\frac{x}{2004}+\frac{x}{1002}+\frac{667x}{668}-2004=0\)
\(\Leftrightarrow\frac{x}{2004}+\frac{x}{1002}+\frac{667x}{668}=2004\)
\(\Rightarrow\frac{x+2x+2001x}{2004}=2004\)
\(\Rightarrow\frac{2004x}{2004}=2004\)
\(\Rightarrow x=2004\)
\(\frac{x}{2004}+\frac{x}{1002}+\frac{667x}{668}-2004=0\)
\(\Leftrightarrow\frac{x}{2004}+\frac{x}{1002}+\frac{667x}{668}=2004\)
\(\Rightarrow\frac{x+2x+2001x}{2004}=2004\)
\(\Rightarrow\frac{2004x}{2004}=2004\)
\(\Rightarrow x=2004\)
Cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)và \(x^2+y^2=1\)
Chứng minh : \(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{2}{\left(a+b\right)^{1002}}\)
cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\)và \(x^2+y^2=1\). CMR: \(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)
Cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) và \(x^2+y^2=1\) CMR : \(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)
Cho \(\frac{x^4}{a}+\frac{y^4}{b}=\frac{1}{a+b}\) và \(x^2+y^2=1\) Chứng minh rằng: \(\frac{x^{2004}}{a^{1002}}+\frac{y^{2004}}{b^{1002}}=\frac{2}{\left(a+b\right)^{102}}\)
\(\frac{x^4}{a}\)=\(\frac{y^4}{b}\)=\(\frac{1}{a+b}\)và x2+y2=1
CMR:\(\frac{x^{2004}}{a^{1002}}\)+\(\frac{y^{2004}}{b^{1002}}\)=\(\frac{2}{\left(a+b\right)^{1002}}\)
Cho \(\frac{x^4}{a}\)+ \(\frac{y^4}{b}\)= \(\frac{1}{a+b}\) và \(x^2\) + \(y^2\)= 1
CMR: \(\frac{x^{2004}}{a^{1002}}\) + \(\frac{y^{2004}}{b^{1002}}\) = \(\frac{2}{\left(a+b\right)^{1002}}\)
Tìm x: a, \(\frac{x-2004}{2003}+\frac{x-2003}{2004}+\frac{x-2005}{2004}=3+\frac{2005}{2003}\)\(+\frac{2004}{2005}\)
a) \(\left(\frac{2}{3}x-\frac{4}{9}\right).\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\)
b) \(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}=-3\)
Tính : P = \(\frac{\frac{1}{2003}+\frac{1}{2004}+\frac{1}{2005}}{\frac{5}{2003}+\frac{5}{2004}-\frac{5}{2005}}-\frac{\frac{2}{2002}+\frac{2}{2003}-\frac{2}{2004}}{\frac{3}{2002}+\frac{3}{2003}-\frac{3}{2004}}\)