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A = 1+(2-3-4+5)+(6-7-8+9)+.....+(2006-2007-2008+2009)+2010
= 1+0+0+.....+0+2010 = 2011
k mk nha
Ta có:
\(\frac{A}{B}=\frac{\frac{2000}{1}+\frac{1999}{2}+\frac{1998}{3}+...+\frac{1}{2000}+2000}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{\left(\frac{2000}{1}+1\right)+\left(\frac{1999}{2}+1\right)+\left(\frac{1998}{3}+1\right)+...+\left(\frac{1}{2000}+1\right)+2000+1}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{\frac{2001}{1}+\frac{2001}{2}+\frac{2001}{3}+...+\frac{2001}{2000}+2001}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=\frac{2001\left(1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}\right)}{1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{2000}}\)
\(\Leftrightarrow\frac{A}{B}=2001\)
bn cộng trên tử rồi thì phải trừ đi chứ ko phân số sẽ thay đổi
\(=\frac{19}{4}+\left(-\frac{37}{100}\right)+\left(-\frac{32}{25}\right)+\left(-\frac{5}{2}\right)+\frac{7}{2}\)
\(=\frac{219}{50}+\left(-\frac{32}{25}\right)+\left(-\frac{5}{2}\right)+\frac{7}{2}\)
\(=\frac{31}{10}+\left(-\frac{5}{2}\right)+\frac{7}{2}\)
\(=\frac{31}{10}+\left(-\frac{25}{10}\right)+\frac{35}{10}\)
\(=\frac{41}{10}\)
\(4\frac{3}{4}+\left(-0,37\right)+\left(-1,28\right)+\left(-2,5\right)+3\frac{1}{2}\)
\(=\frac{19}{4}-\frac{37}{100}-\frac{128}{100}-\frac{25}{10}+\frac{7}{2}\)
\(=\frac{475}{100}-\frac{37}{100}-\frac{128}{100}-\frac{250}{100}+\frac{350}{100}\)
\(=\frac{410}{100}=4,1\)
2B = 2 + 22 + 23 + .... + 22009
2B - B = [2 + 22 + 23 + .... + 22009] - [1 + 2 + 22 + ... + 22008]
B = 22009 - 1
Vậy dãy trên bằng:
\(\frac{2^{2009}-1}{1-2^{2009}}=-1\)
b)
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\)
\(=\frac{1}{1.3}+\frac{1}{2.3}+\frac{1}{2.5}+...+\frac{2}{x.\left(x+1\right)}=\frac{2007}{2009}\)
\(=\frac{1}{2}.\left(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x.\left(x+1\right)}\right)=\frac{2007}{2009}\)
\(=\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2007}{2009}:\frac{1}{2}\)
\(=\frac{1}{2}-\frac{1}{x+1}=\frac{2007}{4018}\)
\(=\frac{1}{x-1}=\frac{1}{2009}\Leftrightarrow x+1=2009\)
\(\Rightarrow x=2009-1=2008\)
Bạn Phúc Trần Tấn bạn có biết làm phần a ko?Giúp mk với ạ!Mai mk cần rùi
a) (-1) + 2 + (-3) + 4 + .... + (-2009) + 2010
= (-1 + 2) + (-3 + 4) + ..... + (-2009 + 2010)
= -1 + (-1) + (-1) + .... + (-1)
= -1 . 1005 = -1005
b) 1 + (-2) + (-3) + 4 + 5 + (-6) + (-7) + 8 + ... + 2005 + (-2006) + (-2007) + 2008
= [1 + (-2) + (-3) + 4] + [5 + (-6) + (-7) + 8 ] + ..... + [2005 + (-2006) + (-2007) + 2008]
= 0 + 0 + ...... + 0 = 0
\(=\frac{2^2-1}{2^2}\cdot\frac{3^2-1}{3^2}\cdot\cdot\cdot\frac{2016^2-1}{2016^2}=\frac{1.3}{2.3}\cdot\frac{2.4}{3.3}\cdot\cdot\cdot\cdot\frac{2015.2017}{2016.2016}\)
\(=\frac{\left(1.2.3....2015\right).\left(3.4....2016.2017\right)}{\left(2.3....2016\right)\left(2.3......2015.2016\right)}=\frac{2017}{2.2016}=\frac{2017}{4032}\)
Đặt A=2+23 +24+...+ 22009
2A=2.(2+23 +24+...+ 22009)=22+24+25+....+22010
A=2A - A=(22+24+25+....+22010) - (2+23 +24+...+ 22009)
=22-2+23+22010=10+22010
Vậy :2+23 +24+...+ 22009=10+22010
Đúng đúng ko đúng thì sai nhé ;)