tính Q biết Q=\(\frac{2\cdot2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+.......+\frac{1}{1+2+3+....+2012}}\)
tính Q
Q=\(\frac{2\cdot2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+......+\frac{1}{1+2+3+....+2012}}\)
\(\frac{2\cdot2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}}\)
\(A=\frac{2.2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}}\)
\(A=\frac{4024}{1+\frac{1}{2.3:2}+\frac{1}{3.4:2}+...+\frac{1}{2012.2013:2}}\)
\(A=\frac{4024}{1+\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{2012.2013}}\)
\(A=\frac{4024}{1+2\left(\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{2012.2013}\right)}\)
\(A=\frac{4024}{1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)}\)
\(A=\frac{4024}{1+2\left(\frac{1}{2}-\frac{1}{2013}\right)}\)
\(A=\frac{4024}{1+1-\frac{2}{2013}}=\frac{4024}{2-\frac{2}{2013}}=4024:\frac{4024}{2013}=\frac{4024.2013}{4024}=2013\)
Tính:
a) M=\(\frac{2\cdot2012}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+20012}}\)
b) N= \(1+\frac{1}{2\cdot\left(1+2\right)}+\frac{1}{3\cdot\left(1+2+3\right)+}+\frac{1}{4}\cdot\left(1+2+3+4\right)\)\(+...+\frac{1}{16}\cdot\left(1+2+3+4+...+16\right)\)
Giúp nha mình tick
\(A=\frac{1\cdot2}{2\cdot2}\cdot\frac{2\cdot3}{3\cdot3}\cdot\frac{3\cdot4}{4\cdot4}\cdot\frac{4\cdot5}{5\cdot5}\cdot.................\cdot\frac{2012\cdot2013}{2013\cdot2013}\)với
\(B=\frac{2012\cdot2013-2012\cdot2012}{2012\cdot2011+2012\cdot2}\)
A=\(\frac{1}{2}\).\(\frac{2}{3}\)....\(\frac{2012}{2013}\)=\(\frac{1}{2013}\)
B=\(\frac{2012}{2012.2013}\)=\(\frac{1}{2013}\)
vậy A=B
\(A=\frac{1.2}{2.2}.\frac{2.3}{3.3}.\frac{3.4}{4.4}.\frac{4.5}{5.5}.....\frac{2012.2013}{2013.2013}=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}....\frac{2012}{2013}=\frac{1.2.3.4.5....2012}{2.3.4.5....2013}=\frac{1}{2013}\)
\(B=\frac{2012.2013-2012.2012}{2012.2011+2012.2}=\frac{2012.\left(2013-2012\right)}{2012.\left(2011+2\right)}=\frac{2012}{2012.2013}=\frac{1}{2013}\)
\(\Rightarrow A=B\)
tính q
q=\(\frac{2\cdot2010}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+....+\frac{1}{1+2+3+....+2012}}\)
\(Q=\frac{2\cdot2010}{1+\frac{1}{1+2}+\frac{1}{1+2+3}+\frac{1}{1+2+3+4}+...+\frac{1}{1+2+3+...+2012}}\)
\(Q=\frac{2\cdot2010}{1+\frac{1}{\frac{(1+2)\cdot2}{2}}+\frac{1}{\frac{(1+3)\cdot3}{2}}+\frac{1}{\frac{(1+4)\cdot4}{2}}+...+\frac{1}{\frac{(1+2012)\cdot2012}{2}}}\)
\(Q=\frac{2\cdot2010}{1+\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{1}{2025078}}\)
\(Q=\frac{2\cdot2010}{1+\frac{2}{6}+\frac{2}{12}+\frac{2}{20}...+\frac{2}{4050156}}\)
\(Q=\frac{2\cdot2010}{1+\frac{2}{2\cdot3}+\frac{2}{3\cdot4}+\frac{2}{4\cdot5}+...+\frac{2}{2012\cdot2013}}\)
\(Q=\frac{2\cdot2010}{1+2\left[\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right]}\)
\(Q=\frac{2\cdot2010}{1+2\left[\frac{1}{2}-\frac{1}{2013}\right]}=\frac{2\cdot2010}{1+\frac{2011}{2013}}=\frac{2\cdot2010}{\frac{4024}{2013}}=\frac{4020}{\frac{4024}{2013}}=4020\cdot\frac{2013}{4024}=...\)
Nguyễn Linh Chi ơi , hình như cô nhầm thì phải :v \(2-\frac{2}{2013}=\frac{2\cdot2013-2}{2013}=\frac{4026-2}{2013}=\frac{4024}{2013}\)
sao mà bằng \(\frac{4020}{2013}\)được cô
Ta có:
\(P=1+\frac{1}{1+2}+\frac{1}{1+2+3}+...+\frac{1}{1+2+3+...+2012}\)
\(P=1+\frac{1}{\frac{\left(1+2\right).2}{2}}+\frac{1}{\frac{\left(1+3\right).3}{2}}+...+\frac{1}{\frac{\left(1+2012\right).2012}{2}}\)
\(P=1+\frac{2}{3.2}+\frac{2}{4.3}+...+\frac{2}{2013.2012}\)
\(P=1+2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{2012}-\frac{1}{2013}\right)\)\
\(P=1+2\left(\frac{1}{2}-\frac{1}{2013}\right)\)
\(P=1+1-\frac{2}{2013}=2-\frac{2}{2013}=\frac{4020}{2013}\)
\(Q=\frac{2.2010}{P}=\frac{4020}{\frac{4020}{2013}}=2013\)....
Tính :
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}}\)
Mẫu số của A \(=\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}\)
\(=\left(1+1+...+1\right)+\left(\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}\right)\)
(2012 số 1) (2011 phân số)
\(=\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{1}{2012}\right)+1\)
\(=\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}\)
\(=2013.\left(\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}\right)\)
=> \(A=\frac{1}{2013}\)
\(A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2012}{1}+\frac{2011}{2}+\frac{2010}{3}+...+\frac{1}{2012}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\left(1+\frac{2011}{2}\right)+\left(1+\frac{2010}{3}\right)+...+\left(1+\frac{1}{2012}\right)+1}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{\frac{2013}{2}+\frac{2013}{3}+...+\frac{2013}{2012}+\frac{2013}{2013}}\)
\(\Rightarrow A=\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2013.\left(\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}\right)}\)
\(\Rightarrow A=\frac{1}{2013}\)
Vậy \(A=\frac{1}{2013}\)
Tính\(\frac{\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2013}}{2012+\frac{2012}{2}+\frac{2011}{3}+\frac{2010}{4}+...+\frac{1}{2013}}\)
\(\frac{\frac{1}{2}+\frac{1}{3}+......+\frac{1}{2013}}{\frac{2012}{1}+\frac{2012}{2}+\frac{2011}{3}+...+\frac{1}{2013}}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+..+\frac{1}{2013}}{\frac{2012}{1}+2+\frac{2012}{2}+1+\frac{2011}{3}+1+...+\frac{1}{2013}+1-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{\frac{2014}{1}+\frac{2014}{2}+...+\frac{2014}{2013}-2014}\)
=\(\frac{\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}}{2014\left(1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{2013}-1\right)}\)
=\(\frac{1}{2014}\)
Tính \(\frac{1}{2\sqrt{1}+1\sqrt{2}}+\frac{1}{3\sqrt{2}+2\sqrt{3}}+\frac{1}{4\sqrt{3}+3\sqrt{4}}+...+\frac{1}{2013\sqrt{2012}+2012\sqrt{2013}}\)
= \(\frac{1}{\sqrt{2}\left(\sqrt{2}+1\right)}+\frac{1}{\sqrt{6}\left(\sqrt{3}+\sqrt{2}\right)}+...+\frac{1}{\sqrt{2012}.\sqrt{2013}\left(\sqrt{2013}+\sqrt{2012}\right)}\)
= \(\frac{\left(\sqrt{2}+1\right)\left(\sqrt{2}-1\right)}{\sqrt{2\left(\sqrt{2}+1\right)}}+...+\frac{\left(\sqrt{2013}-\sqrt{2012}\right)\left(\sqrt{2013}+\sqrt{2012}\right)}{\sqrt{2012}\sqrt{2013}\left(\sqrt{2012}+\sqrt{2013}\right)}\)
= \(\frac{\sqrt{2}-1}{\sqrt{2}}+...+\frac{\sqrt{2013}-\sqrt{2012}}{\sqrt{2012}\sqrt{2013}}\)
= \(1-\frac{1}{\sqrt{2}}+\frac{1}{\sqrt{2}}-\frac{1}{\sqrt{3}}+...+\frac{1}{\sqrt{2012}}-\frac{1}{\sqrt{2013}}\)
= \(\frac{\sqrt{2013}-1}{\sqrt{2013}}=\frac{2013-\sqrt{2013}}{2013}\)