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kết quả là 2008 đấy bạn
nếu nhà bạn có máy tính thì chỉ cần bấm phương trình x thì sẽ ra kết quả thôi
\(\frac{x-1}{2007}+\frac{x-2}{2006}+\frac{x-3}{2005}=\frac{x-4}{2004}+\frac{x-5}{2003}+\frac{x-6}{2002}\)
=> \(\left(\frac{x-1}{2007}-1\right)+\left(\frac{x-2}{2006}-1\right)+\left(\frac{x-3}{2005}-1\right)=\left(\frac{x-4}{2004}-1\right)+\left(\frac{x-5}{2003}-1\right)+\left(\frac{x-6}{2002}-1\right)\)
=> \(\frac{x-1+2007}{2007}+\frac{x-2+2006}{2006}+\frac{x-3+2005}{2005}=\frac{x-4+2004}{2004}+\frac{x-5+2003}{2003}+\frac{x-6+2002}{2002}\)
=> \(\frac{x-2008}{2007}+\frac{x-2008}{2006}+\frac{x-2008}{2005}=\frac{x-2008}{2004}+\frac{x-2008}{2003}+\frac{x-2008}{2002}\)
=> \(\frac{x-2008}{2007}+\frac{x-2008}{2006}+\frac{x-2008}{2005}-\frac{x-2008}{2004}-\frac{x-2008}{2003}-\frac{x-2008}{2002}=0\)
=> \(\left(x-2008\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
Mà \(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\ne0\)
=> x - 2008 = 0 => x = 2008
Vậy x = 2008
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}=\frac{1}{2005}\)
\(\Rightarrow x+1=2005\Rightarrow x=2004\)
Bạn tham khảo nhé
Ta có công thức :
\(\frac{a}{b}< \frac{a+c}{b+c}\) \(\left(\frac{a}{b}< 1;a,b,c\inℕ^∗\right)\)
Áp dụng vào ta có :
\(B=\frac{2004^{2004}+1}{2004^{2005}+1}< \frac{2004^{2004}+1+2003}{2004^{2005}+1+2003}=\frac{2004^{2004}+2004}{2004^{2005}+2004}=\frac{2004\left(2004^{2003}+1\right)}{2004\left(2004^{2004}+1\right)}=\frac{2004^{2003}+1}{2004^{2004}+1}\)
Lại có :
\(A=\frac{2004^{2003}+1}{2004^{2004}+1}\)
\(\Rightarrow\)\(B< A\) hay \(A>B\)
Vậy \(A>B\)
Mình làm được phần a thôi. Sorry
a, Đặt \(A=\frac{1}{2}+\left(\frac{1}{2}\right)^2+...+\left(\frac{1}{2}\right)^{100}\)
\(2A=1+\frac{1}{2}+...+\left(\frac{1}{2}\right)^{99}\)
\(2A-A=\left[1+\frac{1}{2}+...+\left(\frac{1}{2}\right)^{99}\right]-\left[\frac{1}{2}-\left(\frac{1}{2}\right)^2-...-\left(\frac{1}{2}\right)^{100}\right]\)
\(A=1-\left(\frac{1}{2}\right)^{100}\)
Vậy A<2