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\(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}=-3\)
\(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}+3=0\)
\(\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+6}{2004}+1\right)+\left(\frac{x+7}{2003}+1\right)=0\)
\(\frac{x+5+2005}{2005}+\frac{x+6+2004}{2004}+\frac{x+7+2003}{2003}=0\)
\(\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2012}{2003}=0\)
\(\left(x+2010\right)\left(\frac{1}{2005}+\frac{1}{2006}+\frac{1}{2007}\right)=0\)
\(x+2010=0\)
\(x=-2010\)
\(\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=0\)
<=> \(\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)
<=>\(\left(x+2010\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)
<=>x+2010=0
<=>x=-2010
\(\frac{x+1}{1974}+\frac{x+2}{1973}+\frac{x+3}{1972}=-3\)
\(\Rightarrow\left(\frac{x+1}{1974}+1\right)+\left(\frac{x+2}{1973}+1\right)+\left(\frac{x+3}{1972}+1\right)=0\)
\(\Rightarrow\frac{x+1+1974}{1974}+\frac{x+2+1973}{1973}+\frac{x+3+1972}{1972}=0\)
\(\Rightarrow\frac{x+1975}{1974}+\frac{x+1975}{1973}+\frac{x+1975}{1972}=0\)
\(\Rightarrow\left(x+1975\right)\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}=0\)
Mà \(\frac{1}{1974}+\frac{1}{1973}+\frac{1}{1972}\ne0\)
\(\Rightarrow x+1975=0\)
\(\Rightarrow x=0+1975\)
\(\Rightarrow x=1975\)
Vậy \(x=1975\)
b) phần này làm tương tự phần a nha, chuyển -3 sang vế bên trái r cộng từng p.số vs 1 và sau đó nhóm tử số chung ra ngoài ^^
a)\(\left(\frac{2}{3}x-\frac{4}{9}\right).\left(\frac{1}{2}+\frac{-3}{7}:x\right)=0\)
\(\frac{2}{3}x-\frac{4}{9}=0\)hoặc\(\frac{1}{2}+\frac{-3}{7}:x=0\)
\(\frac{2}{3}x=\frac{4}{9}\)hoặc\(-\frac{3}{7}:x=-\frac{1}{2}\)
\(x=\frac{4}{9}:\frac{2}{3}\)hoặc\(x=-\frac{3}{7}:\frac{-1}{2}\)
\(x=\frac{2}{3}\)hoặc\(x=\frac{6}{7}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)
\(=\frac{1}{1}-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)
\(=\frac{1}{1}-\frac{1}{100}=\frac{100}{100}-\frac{1}{100}=\frac{99}{100}\)
\(\frac{3}{1.4}+\frac{3}{4.7}+\frac{3}{7.10}+...+\frac{3}{94.97}+\frac{3}{97.100}\)
\(=1-\frac{1}{4}+\frac{1}{4}-\frac{1}{7}+\frac{1}{7}-\frac{1}{10}+...+\frac{1}{94}-\frac{1}{97}+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}\)
\(=\frac{99}{100}\)
\(pt\Leftrightarrow\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+7}{2003}+1\right)=0\)
\(\Leftrightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\)
Mà \(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\ne0\Rightarrow x+2010=0\Rightarrow x=-2010\)
Vậy \(x=-2010\)
\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-3}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}+1\right)+\left(\dfrac{x-7}{2002}+1\right)+\left(\dfrac{x-6}{2003}+1\right)=\left(\dfrac{x-5}{2004}+1\right)+\left(\dfrac{x-4}{2005}+1\right)+\left(\dfrac{x-3}{2006}+1\right)\)
\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2005}-\dfrac{x-2009}{2006}=0\)
\(\Leftrightarrow\left(x-2009\right).\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)
\(\text{Mà}:\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)\ne0\)
\(\Rightarrow x-2009=0\Rightarrow x=2009\)
\(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}=\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-3=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{4}+\dfrac{x-5}{2006}\right)-3\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}\right)-\left(1+1+1\right)=\left(\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}\right)-\left(1+1+1\right)\)
\(\Leftrightarrow\dfrac{x-8}{2001}+\dfrac{x-7}{2002}+\dfrac{x-6}{2003}-1-1-1=\dfrac{x-5}{2004}+\dfrac{x-4}{2005}+\dfrac{x-5}{2006}-1-1-1\)
\(\Leftrightarrow\left(\dfrac{x-8}{2001}-1\right)+\left(\dfrac{x-7}{2002}-1\right)+\left(\dfrac{x-6}{2003}-1\right)=\left(\dfrac{x-5}{2004}-1\right)+\left(\dfrac{x-4}{2005}-1\right)+\left(\dfrac{x-5}{2006}-1\right)\)
\(\)\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}=\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\)
\(\Leftrightarrow\left(\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}\right)-\left(\dfrac{x-2009}{2004}+\dfrac{x-2009}{2006}+\dfrac{x-2009}{2006}\right)=0\)
\(\Leftrightarrow\dfrac{x-2009}{2001}+\dfrac{x-2009}{2002}+\dfrac{x-2009}{2003}-\dfrac{x-2009}{2004}-\dfrac{x-2009}{2006}-\dfrac{x-2009}{2006}=0\)
\(\Leftrightarrow\left(x-2009\right)\left(\dfrac{1}{2001}+\dfrac{1}{2002}+\dfrac{1}{2003}-\dfrac{1}{2004}-\dfrac{1}{2005}-\dfrac{1}{2006}\right)=0\)
\(\Leftrightarrow x-2009=0\)
\(\Leftrightarrow x=2009\)
Vậy \(x=2009\)
\(\Leftrightarrow\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=0\)
\(\Leftrightarrow\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)
\(\Leftrightarrow\left(x+2010\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\ne0\)
\(\Leftrightarrow x=-2010\)
câu 6 dễ mà câu 7
A=3-3/4+3/4-3/7+3/7-3/10+...+3/94-3/97+3/97-3/100
A=3-3/100
A=300/100-3/100
A=297/100
Câu 6:
\(\frac{x+5}{2005}+\frac{x+6}{2004}+\frac{x+7}{2003}=-3\)
=>\(\left(\frac{x+5}{2005}+1\right)+\left(\frac{x+6}{2004}+1\right)+\left(\frac{x+7}{2003}+1\right)=-3+3=0\)
=>\(\frac{x+2010}{2005}+\frac{x+2010}{2006}+\frac{x+2010}{2007}=0\)
=>x+2010=0
=>x=-2010
Câu 7:
\(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\cdots+\frac{3}{97\cdot100}\)
\(=1-\frac14+\frac14-\frac17+\cdots+\frac{1}{97}-\frac{1}{100}\)
\(=1-\frac{1}{100}=\frac{99}{100}\)