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\(A=\frac{1}{19}+\frac{9}{19.29}+...+\frac{9}{1999.2009}\)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{29}+...+\frac{1}{1999}-\frac{1}{2009}\right)\)

\(=\frac{1}{19}+\frac{9}{10}\left(\frac{1}{19}-\frac{1}{2009}\right)\)

đến đay bn tự tính nha

Áp dụng tc dãy tỉ số băng nhau ta có:

\(\frac{x}{y+z+1}=\frac{y}{z+x+1}=\frac{z}{x+y-2}=\frac{x+y+x}{2\left(x+y+z\right)}=\frac{1}{2}\)

\(\Rightarrow\left\{\begin{matrix}\frac{x}{y+z+1}=\frac{1}{2}\\\frac{y}{z+x+1}=\frac{1}{2}\\\frac{z}{x+y-2}=\frac{1}{2}\\\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}2x=y+z+1\\2y=z+x+1\\2z=x+y-2\end{matrix}\right.\) (1)

Mà \(x+y+z=\frac{1}{2}\Rightarrow\left\{\begin{matrix}y+z=\frac{1}{2}-x\\z+x=\frac{1}{2}-y\\x+y=\frac{1}{2}-z\end{matrix}\right.\) (*)

Thay (*) vào (1) ta được

\(\left\{\begin{matrix}2x=\frac{1}{2}-x+1\\2y=\frac{1}{2}-y+1\\2z=\frac{1}{2}-z-2\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}3x=\frac{3}{2}\\3y=\frac{3}{2}\\3z=\frac{-3}{2}\end{matrix}\right.\)

\(\Rightarrow\left\{\begin{matrix}x=\frac{1}{2}\\y=\frac{1}{2}\\z=-\frac{1}{2}\end{matrix}\right.\)

Vậy...

mk phải công nhận là oOo Chảnh thì sao oOo giỏi nịnh thật.xấu zậy mà bảo đẹp

\(A=1+5+5^2+..+5^{50}\)

\(\Rightarrow5A=5+5^2+5^3+...+5^{51}\)

\(\Rightarrow5A-A=\)\(\left(5+5^2+..+5^{51}\right)-\left(1+5+..+5^{50}\right)\)

\(\Rightarrow4A=5^{51}-1\)

\(\Rightarrow A=\frac{5^{51}-1}{4}\)

\(7^6+7^5-7^4\)

\(=7^4\left(7^2+7-1\right)\)

\(=7^4.55⋮55\left(đpcm\right)\)

\(\Rightarrow\frac{x}{2013}-\left(\frac{2}{20}+\frac{2}{30}+...+\frac{2}{240}\right)=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2013}-2\left(\frac{1}{4.5}+\frac{1}{5.6}+...+\frac{1}{15.16}\right)=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2013}-2\left(\frac{1}{4}-\frac{1}{16}\right)=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2013}-\frac{3}{8}=\frac{5}{8}\)

\(\Rightarrow\frac{x}{2013}=1\)

\(\Rightarrow x=2013\)

Đặt\(c_1=a_1-b_1,c_2=a_2-b_2,c_3=a_3-b_3,c_4=a_4-b_4,c_5=a_5-b_5\)Xét tổng \(c_1+c_2+c_3+c_4+c_5\)

Ta có:\(c_1+c_2+c_3+c_4+c_5\)=\(a_1-b_1+a_2-b_2,+a_3-b_3+a_4-b_4+a_5-b_5=0\)\(\Rightarrow\)Một trong 5 số \(c_1,c_2,c_3,c_4,c_5\) phải có 1 số chẵn

\(\Rightarrow\)\(c_1.c_2.c_3.c_4.c_5⋮2\)

\(\RightarrowĐPCM\)