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ta cs: \(\frac{a+2006}{a-2006}=\frac{b+2005}{b-2005}\)
\(\Rightarrow\frac{a+2006}{b+2005}=\frac{a-2006}{b-2005}=\frac{a}{b}=\frac{2006}{2005}\)
=> dpcm
Đặt
\(\frac{a}{b}=\frac{c}{d}=k\Rightarrow\hept{\begin{cases}a=bk\\c=dk\end{cases}}\)
=> \(\frac{2004a-2005b}{2004a+2005b}=\frac{2004bk-2005b}{2004bk+2005b}=\frac{2004k-2005}{2004k+2005}\left(1\right)\)
\(\frac{2004c-2005d}{2004c+2005d}=\frac{2004dk-2005d}{2004dk+2005d}=\frac{2004k-2005}{2004k+2005}\left(2\right)\)
Từ (1) và (2)
=> \(\frac{2004a-2005b}{2004a+2005b}=\frac{2004c-2005d}{2004c+2005d}\left(đpcm\right)\)
\(\frac{a}{b}< \frac{a+2006}{b+2006}\)
\(\Leftrightarrow a\left(b+2006\right)< b\left(a+2006\right)\)
\(\Leftrightarrow ab+2006a< ab+2006b\)
\(\Leftrightarrow2006a< 2006b\)
\(\Leftrightarrow a< b\) (thỏa mãn đề bài)
Vậy \(\frac{a}{b}< \frac{a+2006}{b+2006}\)
\(1+\frac{1}{3}+\frac{1}{5}+...+\frac{1}{2015}-\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{5}+...+\frac{1}{2015}+\frac{1}{2016}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+...+\frac{1}{2016}\right)\)
\(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2015}+\frac{1}{2016}-\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{1003}\right)\)
\(\frac{1}{1004}+\frac{1}{1005}+...+\frac{1}{2016}\)
Đặt A=1-1/2+1/3-1/4+.......+1/2005-1/2006
=>A= (1+1/3+1/5+...+1/2005)-(1/2+1/4+1/6+.....+1/2006)
=>A=(1+1/2+1/3+...+1/2005)-2.(1/2+1/4+1/6+...+1/2006)
=>A=(1+1/2+1/3+....+1/2005)-(1+1/2+1/3+...+1/1003)
=>A=1/1004+1/1005+.....+1/2006
Vậy A=1/1004+1/1005+.....+1/2006 ( Điều phải chứng minh )
Bài 1:
a) Sửa lại là: \(3^{n+2}-2^{n+2}+3^n-2^n⋮10\) nhé.
\(3^{n+2}-2^{n+2}+3^n-2^n\)
\(=\left(3^{n+2}+3^n\right)-\left(2^{n+2}+2^n\right)\)
\(=3^n.\left(3^2+1\right)-2^n.\left(2^2+1\right)\)
\(=3^n.\left(9+1\right)-2^n.\left(4+1\right)\)
\(=3^n.\left(9+1\right)-2^{n-1}.2.\left(4+1\right)\)
\(=3^n.10-2^{n-1}.2.5\)
\(=3^n.10-2^{n-1}.10\)
\(=10.\left(3^n-2^{n-1}\right)\)
Vì \(10⋮10\) nên \(10.\left(3^n-2^{n-1}\right)⋮10.\)
\(\Rightarrow3^{n+2}-2^{n+2}+3^n-2^n⋮10\left(đpcm\right)\left(\forall n\in N^X\right).\)
Chúc bạn học tốt!