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Số số hạng của B là 1914(là 1 số chẵn)
\(\Rightarrow B=\left(1-\dfrac{1}{2013^2}\right)\left(1-\dfrac{1}{2012^2}\right)\left(1-\dfrac{1}{2011^2}\right)\cdot\cdot\cdot\cdot\cdot\left(1-\dfrac{1}{100^2}\right)\)
\(B=\dfrac{2013^2-1}{2013^2}\cdot\dfrac{2012^2-1}{2012^2}\cdot\dfrac{2011^2-1}{2011^2}\cdot\cdot\cdot\cdot\cdot\dfrac{100^2-1}{100^2}\)
\(B=\dfrac{2014\cdot2012\cdot2013\cdot2011\cdot2012\cdot2010\cdot...\cdot101\cdot99}{2013\cdot2013\cdot2012\cdot2012\cdot2011\cdot2011\cdot...\cdot100\cdot100}\)
\(B=\dfrac{2014\cdot99}{2013\cdot100}=\dfrac{3021}{3050}\)
B=\(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)- \(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}\right)\)
=\(\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{1006}+\dfrac{1}{1007}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)\)-2\(\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
=1-\(\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...-\dfrac{1}{2012}+\dfrac{1}{2013}\)=S
( A-B)2013 =0
Chúc ban học tốt nhé...!
Ta có: \(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(=\left(1+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+...+\dfrac{1}{2011}+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(\Rightarrow P-S=\left(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\right)-\left(\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2013}\right)=0\)
\(\Rightarrow\left(P-S\right)^{2013}=0^{2013}=0\)
Vậy \(\left(P-S\right)^{2013}=0\)
a) Ta có:
\(\frac{x+11}{12}+\frac{x+11}{13}+\frac{x+11}{14}=\frac{x+11}{15}+\frac{x+11}{16}\)
\(\Rightarrow\left(x+11\right)\left(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\right)=\left(x+11\right)\left(\frac{1}{15}+\frac{1}{16}\right)\)
Mà ta có:
\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}\ne\frac{1}{15}+\frac{1}{16}\)
\(\Rightarrow x+11=0\Rightarrow x=-11\)
Ta có:
\(A=1+x+x^2+x^3+...+x^{100}\)
Đặt \(B=x+x^2+x^3+...+x^{100}\)
\(\Rightarrow B=\left(-11\right)+\left(-11\right)^2+\left(-11\right)^3+...+\left(-11\right)^{100}\)
\(\Rightarrow-11B=\left(-11\right)^2+\left(-11\right)^3+\left(-11\right)^4+...+\left(-11\right)^{101}\)
\(\Rightarrow-11B-B=\left(-11\right)^{101}-\left(-11\right)\)
\(\Rightarrow-12B=\left(-11\right)^{101}+11\Rightarrow B=\frac{\left(-11\right)^{101}+11}{-12}\)
\(\Rightarrow A=1+B=\frac{\left(-11\right)^{101}+11}{-12}+1\)
a) \(\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=\sqrt{16}\) \(\Rightarrow\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=4\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}-\dfrac{1}{3}=-2\\\dfrac{x}{2}-\dfrac{1}{3}=2\end{matrix}\right.\) \(\Rightarrow\left[{}\begin{matrix}\dfrac{x}{2}=\dfrac{-5}{3}\\\dfrac{x}{2}=\dfrac{7}{3}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{-10}{3}\\x=\dfrac{14}{3}\end{matrix}\right.\)
Vậy \(x=\dfrac{-10}{3}\) hoặc \(x=\dfrac{14}{3}\) thì thỏa mãn đề bài.
b) \(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\) \(\Rightarrow\left(\dfrac{x+4}{2010}+1\right)+\left(\dfrac{x+3}{2011}+1\right)=\left(\dfrac{x+2}{2012}+1\right)+\left(\dfrac{x+1}{2013}+1\right)\) \(\Rightarrow\dfrac{x+4+2010}{2010}+\dfrac{x+3+2011}{2011}=\dfrac{x+2+2012}{2012}+\dfrac{x+1+2013}{2013}\) \(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\) \(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\) \(\Rightarrow\left(x+2014\right)\times\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\) \(\Rightarrow x+2014=0\) \(\Rightarrow x=-2014\)
Vậy \(x=-2014\) thì thỏa mãn đề bài.
c) \(3^{x+2}+4\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1+1}+4\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1}\times3+4\times3^{x+1}=7\times3^6\) \(\Rightarrow\left(3+4\right)\times3^{x+1}=7\times3^6\) \(\Rightarrow3^{x+1}=3^6\) \(\Rightarrow x+1=6\) \(\Rightarrow x=5\)
Vậy \(x=5\) thì thỏa mãn đề bài.
a)
\(\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=\sqrt{16}\\ \Rightarrow\left(\dfrac{x}{2}-\dfrac{1}{3}\right)^2=4\\ \Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}-\dfrac{1}{3}=2\\\dfrac{x}{2}-\dfrac{1}{3}=-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{1}{3}+2\\\dfrac{x}{2}=\dfrac{1}{3}-2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{2}=\dfrac{7}{3}\\\dfrac{x}{2}=\dfrac{-5}{3}\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{7}{3}.2\\x=\dfrac{-5}{3}.2\end{matrix}\right.\)
\(\Rightarrow\left\{{}\begin{matrix}x=\dfrac{14}{3}\\x=\dfrac{-10}{3}\end{matrix}\right.\)
b)
\(\dfrac{x+4}{2010}+\dfrac{x+3}{2011}=\dfrac{x+2}{2012}+\dfrac{x+1}{2013}\)
\(\Rightarrow\dfrac{x+4}{2010}+1+\dfrac{x+3}{2011}+1=\dfrac{x+2}{2012}+1+\dfrac{x+1}{2013}+1\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}=\dfrac{x+2014}{2012}+\dfrac{x+2014}{2013}\)
\(\Rightarrow\dfrac{x+2014}{2010}+\dfrac{x+2014}{2011}-\dfrac{x+2014}{2012}-\dfrac{x+2014}{2013}=0\)
\(\Rightarrow\left(x+2014\right)\left(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\right)=0\)
mà \(\dfrac{1}{2010}+\dfrac{1}{2011}-\dfrac{1}{2012}-\dfrac{1}{2013}\ne0\)
=> x + 2014 = 0
=> x = -2014
vậy x = -2014
c)\(3^{x+2}+4.3^{x+1}=7.3^6\)
\(\Rightarrow3^{x+1}.3+4.3^{x+1}=7.3^6\\ \Rightarrow3^{x+1}\left(3+4\right)=7.3^6\\ \Rightarrow3^{x+1}.7=7.3^6\\ \Rightarrow3^{x+1}=3^6\\ \Rightarrow x+1=6\\ x=6-1\\ x=5\)
vậy x = 5
c)
Ta có :\(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
\(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{\dfrac{3}{2}}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{2}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{1}{\dfrac{8}{3}}}\) \(=2+\dfrac{1}{1+\dfrac{3}{8}}\) \(=2+\dfrac{1}{\dfrac{11}{8}}\) \(=2+\dfrac{8}{11}\) \(=\dfrac{30}{11}\)
d) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
Ta có :
\(S=1-\dfrac{1}{2}+\dfrac{1}{3}-\dfrac{1}{4}+..........+\dfrac{1}{2011}-\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(=\left(1+\dfrac{1}{3}+..........+\dfrac{1}{2013}\right)-\left(\dfrac{1}{2}+\dfrac{1}{4}+......+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}\right)-2\left(\dfrac{1}{2}+......+\dfrac{1}{2012}\right)\)
\(=\left(1+\dfrac{1}{2}+\dfrac{1}{3}+.....+\dfrac{1}{2013}\right)-\left(1+\dfrac{1}{2}+......+\dfrac{1}{1006}\right)\)
\(=\dfrac{1}{1007}+\dfrac{1}{1008}+..........+\dfrac{1}{2013}\)
\(\Leftrightarrow S-P=\left(\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}\right)-\left(\dfrac{1}{1007}+\dfrac{1}{1008}+....+\dfrac{1}{2013}\right)\)
\(\Leftrightarrow S-P=0\)
\(\Leftrightarrow\left(S-P\right)^{2013}=0^{2013}=0\)
\(1+\dfrac{1}{2}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-2\times\left(\dfrac{1}{2}+\dfrac{1}{4}+...+\dfrac{1}{2010}+\dfrac{1}{2012}\right)\)
\(\Rightarrow1+\dfrac{1}{2}+...+\dfrac{1}{2012}+\dfrac{1}{2013}-\left(1+\dfrac{1}{2}+...+\dfrac{1}{1005}+\dfrac{1}{1006}\right)\)
\(\Rightarrow\dfrac{1}{1007}+\dfrac{1}{1008}+...+\dfrac{1}{2012}+\dfrac{1}{2013}\)
\(\Rightarrow S=P\Rightarrow S-P=0\Rightarrow\left(S-P\right)^{2013}=1\)