Ta có: \(\left(\frac{2006-2005}{2006+2005}\right)^2=\frac{2006^2-2005^2}{2006^2+2005^2}\)
Vậy hai biểu thức trên bằng nhau
Đúng 0
Bình luận (0)
Các câu hỏi tương tự
Cho \(S=\frac{2}{2005+1}+\frac{2^2}{2005^2+1}+...+\frac{2^{n+1}}{2005^{^{2^n}}+1}+...+\frac{2^{2006}}{2006^{2^{2005}}+1}\). So sánh S với \(\frac{1}{1002}\)
Cho S= \(\frac{2}{2005+1}+\frac{2^2}{2005^2+1}+\frac{2^3}{2005^{2^2}+1}+........+\frac{2^{n+1}}{2005^{2^n}+1}+.......+\frac{2^{2006}}{2005^{2^{2006}}+1}\)
So sánh S với \(\frac{1}{1002}\)
Cho \(S=\frac{2}{2005+1}+\frac{2^2}{2005^2+1}+\frac{2^3}{2005^{2^2}+1}+...+\frac{2^{n+1}}{2005^{2^{n+1}}+1}+...+\frac{2^{2006}}{2005^{2^{2006}}+1}\)
So sánh S với \(\frac{1}{1002}\)
Tinh A = \(\frac{\frac{2006}{1}+\frac{2006}{2}+\frac{2006}{3}+........\frac{2006}{2006}+\frac{2006}{2007}}{\frac{1}{2006}+\frac{2}{2005}+\frac{3}{2004}+.........+\frac{2005}{2}+\frac{2006}{1}}\)
so sánh M=\(\frac{2004}{2005}+\frac{2005}{2006}\) và N=\(\frac{2004+2005}{2005+2006}\)
Không lm tính, hãy so sánh: A= \(\frac{2004}{2005}\)+ \(\frac{2005}{2006}\) và B= \(\frac{2004+2005}{2005+2006}\)
so sánh \(\frac{-7}{10^{2005}}+\frac{-15}{10^{2006}}và\frac{-15}{10^{2005}}+\frac{-7}{10^{2006}}\)
Cho S=\(\frac{2}{2005+1}+\frac{2^2}{2005^2+1}+\frac{2^3}{2005^{2^2}}+...\)\(..+\frac{2^{n+1}}{2005^{2^n}}+...+\frac{2^{2006}}{2005^{2^{2005}}+1}\)
So sánh S với \(\frac{1}{1002}\)
So sánh : A = \(\frac{2005^{2005}+1}{2006^{2006}+1}\)và B = \(\frac{2005^{2004}+1}{2005^{2005}+1}\)
GIÚP MÌNH VỚI !!!