tìm tính :\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)......\left(\frac{1}{2004}-1\right)\left(\frac{1}{2005}-1\right)\)
tính tích :
\(\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right).....\left(\frac{1}{2004}-1\right)\left(\frac{1}{2005}-1\right)\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot...\cdot\frac{2004}{2005}\)
\(=\frac{1\cdot2\cdot3\cdot...\cdot2004}{2\cdot3\cdot4\cdot...\cdot2005}\)
\(=\frac{1}{2005}\)
\(=\frac{-1}{2}.\frac{-2}{3}....\frac{-2003}{2004}.\frac{-2004}{2005}\)
\(=\frac{1}{2005}\)
Rút gọn: \(S=\frac{\left(1^4+\frac{1}{4}\right)\left(3^4+\frac{1}{4}\right)\left(5^4+\frac{1}{4}\right)...\left(2004^4+\frac{1}{4}\right)}{\left(2^4+\frac{1}{4}\right)\left(4^4+\frac{1}{4}\right)\left(6^4+\frac{1}{4}\right)...\left(2005^4+\frac{1}{4}\right)}\)
Đề hơi nhầm 1 xíu nhé, 2004 ở dưới và 2005 ở trên :v
1, tính :
\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)..........\left(\frac{1}{2004}-1\right)\left(\frac{1}{2005}-1\right)\)
2, tìm x \(\frac{4x}{2x-\frac{1}{5}}>0\)
\(A=\left(\frac{1}{2}-1\right)\left(\frac{1}{3}-1\right)...\left(\frac{1}{2004}-1\right)\left(\frac{1}{2005}-1\right)\)
\(=\left(-\frac{1}{2}\right)\times\left(-\frac{2}{3}\right)\times...\times\left(-\frac{2003}{2004}\right)\times\left(-\frac{2004}{2005}\right)\)
\(=\frac{1}{2005}\)
***
\(\frac{4x}{2x-\frac{1}{5}}>0\)
\(\Leftrightarrow\begin{cases}4x>0\\2x-\frac{1}{5}>0\end{cases}\)
\(\Leftrightarrow\begin{cases}x>0\\x>\frac{1}{10}\end{cases}\)
\(\Leftrightarrow x>\frac{1}{10}\)
tính :
\(\left(\frac{1}{2}-1\right).\left(\frac{1}{3}-1\right)....\left(\frac{1}{2004}-1\right).\left(\frac{1}{2005}-1\right)\)
tìm x
\(\frac{4x}{2x-\frac{1}{5}}>0\)
Tính tổng : \(f\left(\frac{1}{2005}\right)+f\left(\frac{2}{2005}\right)+.....+\left(\frac{2004}{2005}\right)vớif\left(x\right)=\frac{100^x}{100^x+10}\)
Tính nhanh:
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{5}\right)\times.......\times\left(1-\frac{1}{2003}\right)\times\left(1-\frac{1}{2004}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times....\times\frac{2003}{2004}\)
\(=\frac{1\times2\times3\times...\times2003}{2\times3\times4\times...\times2014}\)
\(=\frac{1}{2014}\)
\(A=\left(1-\frac{1}{2}\right).\left(1-\frac{1}{3}\right).\left(1-\frac{1}{4}\right)...\left(1-\frac{1}{2003}\right).\left(1-\frac{1}{2004}\right)\)
\(A=\left(1-\frac{1}{2}\right)\cdot\left(1-\frac{1}{3}\right)\cdot\left(1-\frac{1}{4}\right)\cdot...\cdot\left(1-\frac{1}{2003}\right)\cdot\left(1-\frac{1}{2004}\right)\)
\(A=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(A=\frac{1\cdot2\cdot3\cdot...\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2003\cdot2004}\)
\(A=\frac{1}{2004}\)
Tính:a)\(\left(\frac{1}{9}-1\right).\left(\frac{1}{10}-1\right)...\left(\frac{1}{2004}-1\right).\left(\frac{1}{2005}-1\right)\)
b)\(81^{10}-27^{13}-9^{21}⋮225\)
a,\(\left(\frac{1}{9}-1\right).\left(\frac{1}{10}-1\right)...\left(\frac{1}{2004}-1\right).\left(\frac{1}{2005}-1\right)\)
\(=\frac{-8}{9}.\frac{-9}{10}...\frac{-2003}{2004}.\frac{-2004}{2005}\)
\(=\frac{\left(-8\right).\left(-9\right)...\left(-2003\right).\left(-2004\right)}{9.10...2004.2005}\)
\(=\frac{-\left(8.9...2003.2004\right)}{9.10...2004.2005}\)
\(=\frac{-8}{2005}\)
b,Ta có: \(81^{10}-27^{13}-9^{21}\)
\(=\left(3^4\right)^{10}-\left(3^3\right)^{13}-\left(3^2\right)^{21}\)
\(=3^{40}-3^{39}-3^{42}\)
\(=3^{39}.3-3^{39}-3^{39}.3^3\)
\(=3^{39}.\left(3-1-3^3\right)\)
\(=3^2.3^{37}.\left(-25\right)\)
\(=3^{37}.\left(-225\right)⋮225\)
Vậy \(81^{10}-27^{13}-9^{21}⋮225\)
Tính nhanh
\(\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times\left(1-\frac{1}{4}\right)\times......\times\left(1-\frac{1}{2003}\right)\times\left(1-\frac{1}{2004}\right)\)
Giúp mk với
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2002}{2003}\cdot\frac{2003}{2004}\)
\(=\frac{1\cdot2\cdot3\cdot....\cdot2002\cdot2003}{2\cdot3\cdot4\cdot5\cdot....\cdot2003\cdot2004}\)
\(=\frac{1}{2004}\)
\(=\frac{1}{2}\cdot\frac{2}{3}\cdot\frac{3}{4}\cdot...\cdot\frac{2003}{2004}=\frac{1\cdot2\cdot3\cdot4....2003}{2\cdot3\cdot4\cdot5....2004}=\frac{1}{2004}\)
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