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#It's the moment when you're in good mood, you accidentally click back =.=
1) Calculate
\(P=1\frac{1}{3}.1\frac{1}{8}.1\frac{1}{15}....1\frac{1}{63}.1\frac{1}{80}\)
\(=\frac{4}{3}.\frac{9}{8}.\frac{16}{15}....\frac{64}{63}.\frac{81}{80}\)
\(=\frac{2.2}{1.3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}....\frac{8.8}{7.9}.\frac{9.9}{8.10}\)
\(=\frac{2.9}{10}=\frac{9}{5}\)
ta có: 10010 + 1 > 10010 - 1
⇒ A = \(\frac{100^{10}+1}{100^{10}-1}< \frac{100^{10}+1-2}{100^{10}-1-2}=\frac{100^{10}-1}{100^{10}-3}=B\)
vậy A < B
kết quả là 2008 đấy bạn
nếu nhà bạn có máy tính thì chỉ cần bấm phương trình x thì sẽ ra kết quả thôi
\(\frac{x-1}{2007}+\frac{x-2}{2006}+\frac{x-3}{2005}=\frac{x-4}{2004}+\frac{x-5}{2003}+\frac{x-6}{2002}\)
=> \(\left(\frac{x-1}{2007}-1\right)+\left(\frac{x-2}{2006}-1\right)+\left(\frac{x-3}{2005}-1\right)=\left(\frac{x-4}{2004}-1\right)+\left(\frac{x-5}{2003}-1\right)+\left(\frac{x-6}{2002}-1\right)\)
=> \(\frac{x-1+2007}{2007}+\frac{x-2+2006}{2006}+\frac{x-3+2005}{2005}=\frac{x-4+2004}{2004}+\frac{x-5+2003}{2003}+\frac{x-6+2002}{2002}\)
=> \(\frac{x-2008}{2007}+\frac{x-2008}{2006}+\frac{x-2008}{2005}=\frac{x-2008}{2004}+\frac{x-2008}{2003}+\frac{x-2008}{2002}\)
=> \(\frac{x-2008}{2007}+\frac{x-2008}{2006}+\frac{x-2008}{2005}-\frac{x-2008}{2004}-\frac{x-2008}{2003}-\frac{x-2008}{2002}=0\)
=> \(\left(x-2008\right)\left(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\right)=0\)
Mà \(\frac{1}{2007}+\frac{1}{2006}+\frac{1}{2005}-\frac{1}{2004}-\frac{1}{2003}-\frac{1}{2002}\ne0\)
=> x - 2008 = 0 => x = 2008
Vậy x = 2008
*)\(\frac{1}{2}+\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+\frac{1}{30}\)
=\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+\frac{1}{5.6}\)
\(=1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}\)
=\(1-\frac{1}{6}\)
=\(\frac{6}{6}-\frac{1}{6}\)
\(=\frac{5}{6}\)
*)\(\frac{2003}{1.2}+\frac{2003}{2.3}+\frac{2003}{3.4}+...+\frac{2003}{2002.2003}\)
\(=\frac{2003}{1}-\frac{2003}{2}+\frac{2003}{2}-\frac{2003}{3}+\frac{2003}{3}-\frac{2003}{4}+...+\frac{2003}{2002}-\frac{2003}{2003}\)
\(=2003-1\)
\(=2002\)
\(\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{10}\)
= \(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}+...+\frac{1}{10}\)
= \(1-\frac{1}{10}\)
= \(\frac{9}{10}\)
Bạn học tốt nhea ♥
1/2003.2002 - 1/2002.2001 - ... - 1/3.2 - 1/2.1
= 1/2003.2002 - (1/2002.2001 + ... + 1/3.2 + 1/2.1)
= 1/2002.2003 - (1/1.2 + 1/2.3 + ... + 1/2001.2002)
= 1/2002.2003 - (1 - 1/2 + 1/2 - 1/3 + ... + 1/2001 - 1/2002)
= 1/2002.2003 - (1 - 1/2002)
= 1/2002.2003 - 2001/2002
= 1/2002.2003 - 2001.2003/2002.2003
= 1-2001.2003/2002.2003