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Biển Cửa Lò, chùa Thiên mụ, núi Ngũ Hành Sơn, chùa Cầu Hội An, kinh thành Huế, đèo Hải Vân
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Ta có:
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2013}\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}+\frac{1}{8}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2012}+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-\frac{1}{4}-...-\frac{1}{1006}\)
\(=\frac{1}{1007}+\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2012}+\frac{1}{2013}\left(1\right)\)
Mà \(P=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\left(2\right)\)
Từ (1) và (2)\(\Rightarrow S=P\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
Vậy...
Mọi người tk mình đi mình đang bị âm nè!!!!!!
Ai tk mình mình tk lại nha !!!
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+.....+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+\frac{1}{6}+.....+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+\frac{1}{6}.....+\frac{1}{2012}\right)\)
\(S=1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+.....+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-........-\frac{1}{1006}\)
\(S=\frac{1}{1007}+\frac{1}{1008}+.......+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}\)
\(\Rightarrow\left(S-P\right)^2=\left(\frac{1}{1007}+\frac{1}{1008}+....+\frac{1}{2012}+\frac{1}{2013}-\frac{1}{1007}-\frac{1}{1008}-....-\frac{1}{2012}-\frac{1}{2013}\right)^2\)
\(\Rightarrow\left(S-P\right)^2=0\)
Vậy \(\left(S-P\right)^2=0\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(S=\left(1+\frac{1}{3}+...+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)+\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(S=\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)
\(S=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2013}\)
\(\Rightarrow S-P=0\)
\(\Rightarrow\left(S-P\right)^{2013}=0\)
a, |3-2x|=x+1
Đặt ĐK x+1>=0
Suy ra 3-2x=\(\orbr{\begin{cases}x+1\\-x-1\end{cases}}\)
TH1:3-2x=x+1
suy ra -3x=-2
suy ra x=\(\frac{2}{3}\)(t/m)
TH2: 3-2x=-x-1
suy ra x=-4(loại vì ktm đk)
vậy x=2/3
b,câu b bản chỉ phân tích vế trái thôi nhé
phân tích 2013 ra 1+1+....+1 ( 2013 số 1 vào tất cả số hag bên trai)
xong bạn dc x=2014
Hok tốt lười giải wa
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+......+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(=\left(1+\frac{1}{3}+......+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+.....+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)-\left(\frac{1}{2}+\frac{1}{4}+....+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+.......+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+.....+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+........+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+......+\frac{1}{2013}\)
\(=P\)
\(\Leftrightarrow S-P=0\)
\(\Leftrightarrow\left(S-P\right)^{2013}=0\)
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(=\left(1+\frac{1}{3}+...+\frac{1}{2011}+\frac{1}{2013}\right)-\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2012}+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=1+\frac{1}{2}+...+\frac{1}{2012}+\frac{1}{2013}-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)
=> S = P => (S - P)2013 = 0
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+\frac{1}{5}-\frac{1}{6}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(\Rightarrow S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1-\frac{1}{2}-\frac{1}{3}-...-\frac{1}{1006}\)
\(\Rightarrow S=\frac{1}{1007}+\frac{1}{1008}+...+\frac{1}{2012}+\frac{1}{2013}\)\(=P\)
\(\Rightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
Tík cho mik nha!
\(S=1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2011}-\frac{1}{2012}+\frac{1}{2013}\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}\right)-2\left(\frac{1}{2}+\frac{1}{4}+...+\frac{1}{2012}\right)\)
\(=\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2011}+\frac{1}{2012}+\frac{1}{2013}\right)-\left(1+\frac{1}{2}+...+\frac{1}{1006}\right)\)
\(=\frac{1}{1007}+\frac{1}{1008}+\frac{1}{1009}+...+\frac{1}{2013}=P\)
=> S - P = 0
Khi đó (S - P)2013 = 02013 = 0