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\(\Leftrightarrow\left(\dfrac{x-1}{2009}-1\right)+\left(\dfrac{x-2}{2008}-1\right)=\left(\dfrac{x-3}{2007}-1\right)+\left(\dfrac{x-4}{2006}-1\right)\)
=>x-2010=0
hay x=2010
Câu 1.8: Giải
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}>\dfrac{1}{2.3}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}>\dfrac{1}{3.4}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{9.10}\)
\(\Rightarrow A>\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{9.10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+...+\dfrac{1}{9}-\dfrac{1}{10}\)
\(A>\dfrac{1}{2}-\dfrac{1}{10}\)
\(A>\dfrac{2}{5}\) (1)
*Ta có: \(\dfrac{1}{2^2}=\dfrac{1}{2.2}< \dfrac{1}{1.2}\)
\(\dfrac{1}{3^2}=\dfrac{1}{3.3}< \dfrac{1}{2.3}\)
...
\(\dfrac{1}{9^2}=\dfrac{1}{9.9}< \dfrac{1}{8.9}\)
\(\Rightarrow A< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{8.9}\)
\(A< 1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{8}-\dfrac{1}{9}\)
\(A< 1-\dfrac{1}{9}\)
\(A< \dfrac{8}{9}\) (2)
Từ (1) và (2) \(\Rightarrow\dfrac{2}{5}< A< \dfrac{8}{9}\)
\(A=\dfrac{2}{20}+\dfrac{2}{30}+\dfrac{2}{42}+...+\dfrac{2}{240}=2\times\left(\dfrac{1}{20}+\dfrac{1}{30}+\dfrac{1}{42}+\dfrac{1}{240}\right)\)
\(A=2\times\left(\dfrac{1}{4\times5}+\dfrac{1}{5\times6}+\dfrac{1}{6\times7}+....+\dfrac{1}{15\times16}\right)\)
\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{6}+\dfrac{1}{6}-\dfrac{1}{7}+...+\dfrac{1}{15}-\dfrac{1}{16}\right)\)
\(A=2\times\left(\dfrac{1}{4}-\dfrac{1}{16}\right)=\dfrac{3}{8}\)
b) cậu đi tìm số sốm hạng là : \(\left(2010-1\right):1+1=2010\)
\(\Rightarrow\)số cặp trong phép tính là : \(2010:2=1005\)(cặp)
\(\Rightarrow B=1-2+3-4+...+2009-2010\)(1005 cặp)
\(\Rightarrow\left(1-2\right)+\left(3-4\right)+...+\left(2009-2010\right)\)
\(\Rightarrow B=\left(-1\right)+\left(-1\right)+...+\left(-1\right)\)(1005 số -1)
\(\Rightarrow B=\left(-1\right).1005\)
\(\Rightarrow B=\left(-1005\right)\)
cậu tik cho mik nhé!!!
a)\(\frac{5}{2}-3\left(\frac{1}{3}-x\right)=\frac{1}{4}-7x\)
\(\Leftrightarrow\frac{5}{2}-1+x=\frac{1}{4}-7x\)
\(\Leftrightarrow8x=-\frac{5}{4}\)
\(\Leftrightarrow x=-\frac{5}{32}\)
c)\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+...+\frac{2}{x\left(x+1\right)}=\frac{2001}{2003}\)
\(\Leftrightarrow2\left(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+...+\frac{1}{x\left(x+1\right)}\right)=\frac{2001}{2003}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+...+\frac{1}{x\left(x+1\right)}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+...+\frac{1}{x}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2001}{4006}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2003}\)
\(\Leftrightarrow x+1=2003\)
\(\Leftrightarrow x=2002\)
Đặt \(A=\dfrac{2009^{2008}+1}{2009^{2009}+1}\) và \(B=\dfrac{2009^{2007}+1}{2009^{2008}+1}\)
Ta có:
\(2009A=\dfrac{2009.\left(2009^{2008}+1\right)}{2009^{2009}+1}=\dfrac{2009^{2009}+2009}{2009^{2009}+1}\)
\(=\dfrac{2009^{2009}+1+2008}{2009^{2009}+1}=\dfrac{2009^{2009}+1}{2009^{2009}+1}+\dfrac{2008}{2009^{2009}+1}\)
\(=1+\dfrac{1}{2009^{2009}+1}\)
\(2009B=\dfrac{2009.\left(2009^{2007}+1\right)}{2009^{2008}+1}=\dfrac{2009^{2008}+2009}{2009^{2008}+1}\)
\(=\dfrac{2008^{2008}+1+2008}{2009^{2008}+1}=\dfrac{2008^{2008}+1}{2009^{2008}+1}+\dfrac{2008}{2009^{2008}+1}\)
\(=1+\dfrac{2008}{2009^{2008}+1}\)
Vì \(1+\dfrac{2008}{2009^{2009}+1}< 1+\dfrac{2008}{2009^{2008}+1}\)
Nên \(10A< 10B\) \(\Rightarrow A< B\)
Vậy \(\dfrac{2009^{2008}+1}{2009^{2009}+1}< \dfrac{2009^{2007}+1}{2009^{2008}+1}\)
~ Học tốt ~
Nếu:
\(\dfrac{a}{b}< 1\Rightarrow\dfrac{a+m}{b+m}< 1\left(m\in N\right)\)
\(A=\dfrac{2009^{2008}+1}{2009^{2009}+1}< 1\)
\(\Rightarrow A< \dfrac{2009^{2008}+1+2008}{2009^{2009}+1+2008}\Rightarrow A< \dfrac{2009^{2008}+2009}{2009^{2009}+2009}\Rightarrow A< \dfrac{2009\left(2009^{2007}+1\right)}{2009\left(2009^{2008}+1\right)}\Rightarrow A< \dfrac{2009^{2007}+1}{2009^{2008}+1}=B\)\(\Rightarrow A< B\)
Câu1:
a: \(=2008^2-\left(2008-2\right)\left(2008+2\right)\)
\(=2008^2-\left(2008^2-4\right)\)
=4
b: \(=\dfrac{23\cdot29\cdot10101}{23\cdot29\cdot10101}=1\)
c: \(=\dfrac{\left(2^{17}+5^{17}\right)\left(3^{14}-5^{12}\right)\cdot\left(16-16\right)}{15^2+5^3+67^7}\)
=0
Đặt D1 = \(\dfrac{2010}{1}\) + \(\dfrac{2009}{2}\) + \(\dfrac{2008}{3}\) + ... + \(\dfrac{1}{2010}\)
= 1 + ( 1+ \(\dfrac{2009}{2}\)) + ( 1+ \(\dfrac{2008}{3}\)) + ... + (1+\(\dfrac{1}{2010}\))
= \(\dfrac{2011}{2}\) + \(\dfrac{2011}{3}\)+ ... + \(\dfrac{2011}{2010}\) + \(\dfrac{2011}{2011}\)
= 2011. ( \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\))
Đặt D2 = \(\dfrac{1}{2}\) + \(\dfrac{1}{3}\) + ... + \(\dfrac{1}{2010}\) + \(\dfrac{1}{2011}\)
=> D = 2011
cho mk 1 tick nha 
Ta có:
\(2007A=\dfrac{2007^{2009}+2007}{2007^{2009}+1}=1+\dfrac{2006}{2007^{2009}+1}\)\(2007B=\dfrac{2007^{2010}+10}{2007^{2010}+1}=1+\dfrac{9}{2007^{2010}+1}\)Vì \(\dfrac{2007}{2007^{2009}+1}>\dfrac{2007}{2007^{2010}+1}\)
=>2007A > 2007B
Do đó A>B
Vậy A>B
Ta có : \(B\) = \(\dfrac{2007^{2009}+1}{2007^{2010}+1}\) \(< 1\) \(\Rightarrow\dfrac{2007^{2009}+1}{2007^{2010}+1}< \dfrac{2007^{2009}+1+2006}{2007^{2010}+1+2006}\) \(=\dfrac{2007^{2009}+2007}{2007^{2010}+2007}\)
\(=\dfrac{2007\left(2007^{2008}+1\right)}{2007\left(2007^{2009}+1\right)}\) \(=\dfrac{2007^{2008}+1}{2007^{2009}+1}=A\)
Vậy \(A>B\)
Ta có :
\(A=\dfrac{\dfrac{2008}{1}+\dfrac{2007}{2}+....................+\dfrac{2}{2007}+\dfrac{1}{2008}}{\dfrac{1}{2}+\dfrac{1}{3}+....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\left(\dfrac{2007}{2}+1\right)+.....+\left(\dfrac{2}{2007}+1\right)+\left(\dfrac{1}{2008}+1\right)+1}{\dfrac{1}{2}+\dfrac{1}{3}+...............+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{\dfrac{2009}{2}+...................+\dfrac{2009}{2007}+\dfrac{2009}{2008}+\dfrac{2009}{2009}}{\dfrac{1}{2}+\dfrac{1}{3}+.....................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=\dfrac{2009\left(\dfrac{1}{2}+..........................+\dfrac{1}{2008}+\dfrac{1}{2009}\right)}{\dfrac{1}{2}+\dfrac{1}{3}+............................+\dfrac{1}{2008}+\dfrac{1}{2009}}\)
\(\Rightarrow A=2009\)