Mấy chế zô giải giùm tui câu này:
Cho \(A=\left(\dfrac{1}{2^2}-1\right)\left(\dfrac{1}{3^2}-1\right)\left(\dfrac{1}{4^2}-1\right).....\left(\dfrac{1}{2013^2}-1\right)\left(\dfrac{1}{2014^2}-1\right)\) và \(B=-\dfrac{1}{2}.\)
Hãy so sánh A và B.
Cho A=\(\left(\dfrac{1}{2^2}-1\right)\)\(\left(\dfrac{1}{3^2}-1\right)\)\(\left(\dfrac{1}{4^2}-1\right)\)...\(\left(\dfrac{1}{2013^2}-1\right)\)\(\left(\dfrac{1}{2014^2}-1\right)\) và B= \(-\dfrac{1}{2}\)
Hãy so sánh A và B
\(A=-\left(1-\dfrac{1}{2^2}\right)\left(1-\dfrac{1}{3^2}\right)...\left(1-\dfrac{1}{2014^2}\right)\)
\(A=\dfrac{\left(1\cdot3\right)\left(2\cdot4\right)\left(3\cdot5\right)...\left(2012\cdot2014\right)\left(2013\cdot2015\right)}{\left(2\cdot2\right)\left(3\cdot3\right)\left(4\cdot4\right)...\left(2013\cdot2013\right)\left(2014\cdot2014\right)}\)
\(A=\dfrac{\left(1\cdot2\cdot3\cdot...\cdot2012\cdot2013\right)\left(3\cdot4\cdot5\cdot...\cdot2014\cdot2015\right)}{\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)\left(2\cdot3\cdot4\cdot...\cdot2013\cdot2014\right)}\)
\(A=\dfrac{1\cdot2015}{2014\cdot2}=\dfrac{2015}{4028}\)
Vì \(\dfrac{2015}{4028}>-\dfrac{1}{2}\) nên A > B
Tính
\(S=\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{2013}\left(1+2+3+...+2013\right)+\dfrac{1}{2014}\left(1+2+3+...+2014\right)\)
Thực hiện phép tính
a) A= \(1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)\)\(+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{2013}\left(1+2+...+2013\right)\)
b) B=\(\dfrac{1-3}{1.3}+\dfrac{2-4}{2.4}+\dfrac{3-5}{3.5}+\dfrac{4-6}{4.6}+...+\dfrac{2011-2013}{2011.2013}+\dfrac{2012-2014}{2012.2014}-\dfrac{2013+2014}{2013.2014}\)
thực hiện phép tính:
\(\dfrac{1}{x\left(x+1\right)}\)+\(\dfrac{1}{\left(x+1\right)\left(x+2\right)}\)+\(\dfrac{1}{\left(x+2\right)\left(x+3\right)}\)+...+\(\dfrac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)
=1/x-1/x+2014
\(=\dfrac{x+2014-x}{x\left(x+2014\right)}=\dfrac{2014}{x\left(x+2014\right)}\)
Tính:
a) \(A=1+\dfrac{1}{2}\left(1+2\right)+\dfrac{1}{3}\left(1+2+3\right)+\dfrac{1}{4}\left(1+2+3+4\right)+...+\dfrac{1}{2013}\left(1+2+...+2013\right)\)b) \(B=\dfrac{1-3}{1\cdot3}+\dfrac{2-4}{2\cdot4}+\dfrac{3-5}{3\cdot5}+\dfrac{4-6}{4\cdot6}+...+\dfrac{2011-2013}{2011\cdot2013}+\dfrac{2012-2014}{2012\cdot2014}+\dfrac{2013-2015}{2013\cdot2015}\)Giúp mình với!
\(A=1+\dfrac{\dfrac{\left(1+2\right).2}{2}}{2}+\dfrac{\dfrac{\left(1+3\right).3}{2}}{3}+...+\dfrac{\dfrac{\left(1+2013\right).2013}{2}}{2013}\)
\(A=1+\dfrac{\dfrac{3.2}{2}}{2}+\dfrac{\dfrac{4.3}{2}}{3}+...+\dfrac{\dfrac{2014.2013}{2}}{2013}\)
\(A=1+\dfrac{3}{2}+\dfrac{2.3}{3}+...+\dfrac{1007.2013}{2013}\)
\(A=1+\dfrac{3}{2}+2+\dfrac{5}{2}...+1007\)
\(2A=2+3+4+5+6+...+2012+2013+2014\)
\(2A=\dfrac{\left(2+2014\right).2013}{2}\)
\(A=\dfrac{2016.2013}{4}=504.2013\)
\(B=\dfrac{-2}{1.3}+\dfrac{-2}{2.4}+...+\dfrac{-2}{2012.2014}+\dfrac{-2}{2013.2015}\)
\(-B=\dfrac{2}{1.3}+\dfrac{2}{2.4}+...+\dfrac{2}{2012.2014}+\dfrac{2}{2013.2015}\)
\(-B=\left(\dfrac{2}{1.3}+\dfrac{2}{3.5}+...+\dfrac{2}{2013.2015}\right)+\left(\dfrac{2}{2.4}+\dfrac{2}{4.6}+...+\dfrac{2}{2012.2014}\right)\)
\(-B=\left(\dfrac{3-1}{1.3}+\dfrac{5-3}{3.5}+...+\dfrac{2015-2013}{2013.2015}\right)+\left(\dfrac{4-2}{2.4}+\dfrac{6-4}{4.6}+...+\dfrac{2014-2012}{2012.2014}\right)\)
\(-B=\left(1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{2013}-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{4}+\dfrac{1}{4}+...+\dfrac{1}{2012}-\dfrac{1}{2014}\right)\)
\(-B=\left(1-\dfrac{1}{2015}\right)+\left(\dfrac{1}{2}-\dfrac{1}{2014}\right)\)
\(-B=\dfrac{2014}{2015}+\dfrac{2012}{2014.2}=\dfrac{2014^2+1006.2015}{2015.2014}\)
\(B=\dfrac{2014^2+1006.2015}{-2015.2014}\)
Tính
\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+.....+\dfrac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+...+\dfrac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(\Rightarrow A=\dfrac{1}{x}-\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)
\(\Rightarrow A=\dfrac{1}{x}-\dfrac{1}{x+2014}\)
\(\Rightarrow A=\dfrac{2014}{x\left(x+2014\right)}\)
\(A=\dfrac{1}{x\left(x+1\right)}+\dfrac{1}{\left(x+1\right)\left(x+2\right)}+\dfrac{1}{\left(x+2\right)\left(x+3\right)}+....+\dfrac{1}{\left(x+2013\right)\left(x+2014\right)}\)
\(=\dfrac{1}{x}+\dfrac{1}{x+1}+\dfrac{1}{x+1}-\dfrac{1}{x+2}+\dfrac{1}{x+2}-\dfrac{1}{x+3}+...+\dfrac{1}{x+2013}-\dfrac{1}{x+2014}\)
\(=\dfrac{1}{x}-\dfrac{1}{x+2014}-\dfrac{x+2014}{x\left(x+2014\right)}-\dfrac{x}{x\left(x+2014\right)}\)
\(=\dfrac{x+2014-x}{x\left(x+2014\right)}\)
\(=\dfrac{2014}{x\left(x+2014\right)}\)
Câu hỏi today của CLB Yêu Toán❤❤
+ Dễ: \(\left|\dfrac{1}{2}+x\right|-1=\dfrac{11}{2}\); \(1mm=?km\)
+ Khó: \(\left(\dfrac{1}{2}-1\right).\left(\dfrac{1}{3}-1\right).\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2014}-1\right)\)
Bonus: Số nghịch đảo của 50% ?
a)\(\left|\dfrac{1}{2}+x\right|-1=\dfrac{11}{2}\)
\(\Rightarrow\left|\dfrac{1}{2}+x\right|=\dfrac{11}{2}+1=\dfrac{13}{2}\)
\(\Rightarrow\left[{}\begin{matrix}\dfrac{1}{2}+x=\dfrac{-13}{2}\\\dfrac{1}{2}+x=\dfrac{13}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=-7\\x=6\end{matrix}\right.\)
b)\(\left(\dfrac{1}{2}-1\right)\left(\dfrac{1}{3}-1\right)\left(\dfrac{1}{4}-1\right)...\left(\dfrac{1}{2014}-1\right)\)
\(=\dfrac{-1}{2}.\dfrac{2}{-3}.\dfrac{-3}{4}...\dfrac{2012}{-2013}.\dfrac{-2013}{2014}\)
\(=\dfrac{-1}{2014}\)
số nghịch đảo của 50% là:\(\dfrac{100}{50}=2\)
Giải:
Dễ:
|1/2+x|-1=11/2
|1/2+x| =11/2+1
|1/2+x| =13/2
⇒1/2+x=13/2 hoặc 1/2+x=-13/2
x=6 hoặc x=-7
1mm=0,000001km
Khó:
(1/2-1).(1/3-1).(1/4-1).....(1/2014-1)
=-1/2.-2/3.-3/4.....-2013/2014
=-1/2014
Bonus: Số nghịch đảo của 50% là 100/50 =2
tính
a) \(\left[\dfrac{0.8\div\left(\dfrac{4}{5}\cdot1025\right)}{0.64-1}+\dfrac{\left(1.08-\dfrac{2}{25}\right)\div\dfrac{4}{7}}{\left(6\dfrac{5}{7}-3\dfrac{1}{4}\right)\cdot2\dfrac{2}{17}}+\left(1.2\cdot0.5\right)\div\dfrac{4}{5}\right]\)
b) \(\left(0.2\right)^{-3}\left[\left(-\dfrac{1}{5}\right)^{-2}\right]^{-1}+\left[\left(\dfrac{1}{2}\right)^{-3}\right]^{-2}\div\left(2^{-3}\right)^{-1}-\left(0.175\right)^{-2}\)
c) \(2+\dfrac{1}{1+\dfrac{1}{2+\dfrac{1}{1+\dfrac{1}{2}}}}\)
d) \(\dfrac{1}{90}-\dfrac{1}{72}-\dfrac{1}{56}-\dfrac{1}{42}-\dfrac{1}{3}\)
e) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2\div2\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}\cdot\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
g) \(\dfrac{1}{-\left(2017\right)\left(-2015\right)}+\dfrac{1}{\left(-2015\right)\left(-2013\right)}+...+\dfrac{1}{\left(-3\right)\cdot\left(-1\right)}\)
h) \(\left(1-\dfrac{1}{1\cdot2}\right)+\left(1-\dfrac{1}{2\cdot3}+...+\left(1-\dfrac{1}{2017\cdot2018}\right)\right)\)
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) \(\left(\dfrac{1}{3}\right)^{-1}-\left(-\dfrac{6}{7}\right)^0+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\left(\dfrac{1}{2}\right)^2:2\)
\(=3-1+\dfrac{1}{4}:2\)
\(=3-1+\dfrac{1}{8}\)
\(=\dfrac{17}{8}\)
f) \(\dfrac{\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}}{\dfrac{2}{3}-\dfrac{2}{7}-\dfrac{2}{13}}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}{2\left(\dfrac{1}{3}-\dfrac{1}{7}-\dfrac{1}{13}\right)}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}-\dfrac{3}{16}-\dfrac{3}{64}-\dfrac{3}{256}}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{\dfrac{3}{4}\left(1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}\right)}{1-\dfrac{1}{4}-\dfrac{1}{16}-\dfrac{1}{64}}+\dfrac{5}{8}\)
\(=\dfrac{1}{2}.\dfrac{3}{4}+\dfrac{5}{8}\)
\(=\dfrac{3}{8}+\dfrac{5}{8}\)
\(=1\)
\(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right)\left(1\dfrac{2}{7}\right).......\left(1\dfrac{2}{2011}\right)\left(1\dfrac{2}{2013}\right)\)
Ta có : \(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right).....\left(1\dfrac{2}{2011}\right)\left(1\dfrac{2}{2013}\right)\)
\(=\dfrac{5}{3}.\dfrac{7}{5}....\dfrac{2013}{2011}.\dfrac{2015}{2013}=\dfrac{2015}{3}\)
\(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right)\left(1\dfrac{2}{7}\right)...\left(1\dfrac{2}{2011}\right)\left(1\dfrac{2}{2013}\right)\)
\(=\dfrac{5}{3}.\dfrac{7}{5}.\dfrac{9}{7}.....\dfrac{2013}{2011}.\dfrac{2015}{2013}\)
\(=\dfrac{2015}{3}\)
\(\left(1\dfrac{2}{3}\right)\left(1\dfrac{2}{5}\right)\left(1\dfrac{2}{7}\right)...\left(1\dfrac{2}{2013}\right)\)
\(=\dfrac{5}{3}.\dfrac{7}{5}.\dfrac{9}{7}...\dfrac{2015}{2013}=\dfrac{2015}{3}\)