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Bạn chuyển về 1 vế sau đó trừ 1 vào mỗi phân thức ta được :
\(\left(x-2005\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005}\right)=0\)
Vì biểu thức bên phải khác 0 nên : \(x-2005=0\)=> \(x=2005\)
\(\frac{x-5}{2000}+\frac{x-4}{2001}+\frac{x-3}{2002}=\frac{x-2}{2003}+\frac{x-1}{2004}+\frac{x}{2005}\)
\(\Leftrightarrow\frac{x-2005}{2000}+\frac{x-2005}{2001}+\frac{x-2005}{2002}=\frac{x-2005}{2003}+\frac{x-2005}{2004}+\frac{x-2005}{2005}\)
\(\Leftrightarrow\left(x-2005\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}-\frac{1}{2003}-\frac{1}{2004}-\frac{1}{2005}\right)=0\)
<=> x - 2005 = 0
<=> x = 2005
Vậy ...............
\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}\) + 35 = \(^{2^5}\)
\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}\) = -3
\(\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+3}{2002}+1\right)\) = 0
\(\left(\frac{x+1}{2004}+\frac{2004}{2004}\right)+\left(\frac{x+2}{2003}+\frac{2003}{2003}\right)+\left(\frac{x+3}{2002}+\frac{2002}{2002}\right)\)= 0
\(\left(\frac{x+2005}{2004}\right)+\left(\frac{x+2005}{2003}\right)+\left(\frac{x+2005}{2002}\right)\)= 0
\(\left(x+2005\right).\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)\) = 0
\(\left(x+2005\right)\) = 0 \(:\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)\)
\(\left(x+2005\right)\) = 0
\(x\) = 0-2005
\(x\) = -2005
=1+1/2001+1+1/2002+1+1/2003+...+1+1/2008=8+1/2001+1/2002+1/2003+...+1/2008>8
\(\frac{2002}{2001}+\frac{2003}{2002}+\frac{2004}{2003}+\frac{2005}{2004}+\frac{2006}{2005}+\frac{2007}{2006}+\frac{2008}{2007}+\frac{2009}{2008}>8\)
Ta có :
\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5\)
\(\Leftrightarrow\)\(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35\)
\(\Leftrightarrow\)\(\left(\frac{x+1}{2004}+1\right)+\left(\frac{x+2}{2003}+1\right)+\left(\frac{x+3}{2002}+1\right)=32-35+3\)
\(\Leftrightarrow\)\(\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=-3+3\)
\(\Leftrightarrow\)\(\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0\)
Vì \(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0\)
Nên \(x+2005=0\)
\(\Rightarrow\)\(x=-2005\)
Vậy \(x=-2005\)
Chúc bạn học tốt ~
Ta có: \(\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}+35=2^5\)
\(\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=2^5-35\)
\(\Rightarrow\frac{x+1}{2004}+\frac{x+2}{2003}+\frac{x+3}{2002}=-3\)
\(\Rightarrow\frac{x+1}{2004}+1+\frac{x+2}{2003}+1+\frac{x+3}{2002}+1=-3+3\)
\(\Rightarrow\frac{x+1+2004}{2004}+\frac{x+2+2003}{2003}+\frac{x+3+2002}{2002}=0\)
\(\Rightarrow\frac{x+2005}{2004}+\frac{x+2005}{2003}+\frac{x+2005}{2002}=0\)
\(\Rightarrow\left(x+2005\right)\left(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\right)=0\)
Vì \(\frac{1}{2004}+\frac{1}{2003}+\frac{1}{2002}\ne0\)
Nên x + 2005 = 0
=> x = -2005
Vậy x = -2005
a)\(\frac{x+32}{11}+\frac{x+23}{12}=\frac{x+38}{13}+\frac{x+27}{14}\)
\(\left(\frac{x-1}{11}+3\right)+\left(\frac{x-1}{12}+2\right)=\left(\frac{x-1}{13}+3\right)+\left(\frac{x-1}{14}+2\right)\)
\(\left(\frac{x-1}{11}+\frac{x-1}{12}\right)+\left(3+2\right)=\left(\frac{x-1}{13}+\frac{x-1}{14}\right)+\left(3+2\right)\)
\(\frac{x-1}{11}+\frac{x-1}{12}=\frac{x-1}{13}+\frac{x-1}{14}\)
\(\frac{x-1}{11}+\frac{x-1}{12}-\frac{x-1}{13}+\frac{x-1}{14}=0\)
\(\left(x-1\right)\left(\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\right)=0\)
Vì \(\frac{1}{11}+\frac{1}{12}\ne\frac{1}{13}+\frac{1}{14}\)\(\Rightarrow\frac{1}{11}+\frac{1}{12}-\frac{1}{13}+\frac{1}{14}\ne0\)
\(\Rightarrow x-1=0\)
\(\Rightarrow x=1\)
a)\(\frac{2}{6}+\frac{2}{12}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(\frac{2}{2.3}+\frac{2}{3.4}+...+\frac{2}{x\left(x+1\right)}=\frac{2}{2013}\)
\(2\left(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{x}-\frac{1}{x+1}\right)=\frac{2}{2013}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{1}{2013}\)
đề sai
b)\(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(x+2004=0\).Do \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(x=-2004\)
c)\(\frac{x+5}{205}-1+\frac{x+4}{204}-1+\frac{x+3}{203}-1=\frac{x+166}{366}-1+\frac{x+167}{367}-1+\frac{x+168}{368}-1\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}=\frac{x-200}{366}+\frac{x-200}{367}+\frac{x-200}{368}\)
\(\frac{x-200}{205}+\frac{x-200}{204}+\frac{x-200}{203}-\frac{x-200}{366}-\frac{x-200}{367}-\frac{x-200}{368}=0\)
\(\left(x-200\right)\left(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\right)=0\)
\(x-200=0\).Do\(\frac{1}{205}+\frac{1}{204}+\frac{1}{203}-\frac{1}{366}-\frac{1}{367}-\frac{1}{368}\ne0\)
\(x=200\)
d)chịu
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.....+\frac{2}{x\left(x+1\right)}=\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{2}\left(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+....+\frac{2}{x\left(x+1\right)}\right)=\frac{1}{2}.\frac{2003}{2005}\)
\(\Leftrightarrow\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+....+\frac{1}{x\left(x+1\right)}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{x}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{2}-\frac{1}{x+1}=\frac{2003}{4010}\)
\(\Leftrightarrow\frac{1}{x+1}=\frac{1}{2}-\frac{2003}{4010}=\frac{1}{2005}\)
\(\Rightarrow x+1=2005\Rightarrow x=2004\)
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