\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\\ \Rightarrow\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}+3=0\\ \Rightarrow\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+7}{2003}+1\right)=0\\ \Rightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\\ \Rightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\\ \Rightarrow x+2010=0\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\ne0\right)\\ \Rightarrow x=-2010\)

\(\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=-3+3=0\)

\(\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)

\(\left(x+2010\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)

=> x + 2010 = 0

=> x = -2010 

\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\)

=>\(\left(\dfrac{x+5}{2005}+1\right)+\left(\dfrac{x+6}{2004}+1\right)+\left(\dfrac{x+7}{2003}+1\right)=0\)

=>\(\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\)

=>\(x+2010=0\)(do\(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\)khác 0)

=>x=-2010

Vậy...

Chị Nhung ơi , có phải chị chơi Bang Bang k ???

\(\frac{x+5}{2005}+1+\frac{x+6}{2004}+1+\frac{x+7}{2003}+1=0\)

<=> \(\frac{x+2010}{2005}+\frac{x+2010}{2004}+\frac{x+2010}{2003}=0\)

<=>\(\left(x+2010\right)\left(\frac{1}{2005}+\frac{1}{2004}+\frac{1}{2003}\right)=0\)

<=>x+2010=0

<=>x=-2010

Thái Hồ làm đúng rồi nhé Ngọc Vĩ . Bạn đó chuyển sang VT thành +3 rồi tách thành + 1 +1 +1 đó bạn. Bài của Ngọc Vĩ sai rồi

ai ra đề cho 1 lạy

HELP ME!

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:

\(A=\frac{3}{1\cdot4}+\frac{3}{4\cdot7}+\cdots+\frac{3}{97\cdot100}\)

\(=1-\frac14+\frac14-\frac17+\cdots+\frac{1}{97}-\frac{1}{100}\)

\(=1-\frac{1}{100}=\frac{99}{100}\)

Giải:

\(\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}=-3\)

\(\Leftrightarrow\dfrac{x+5}{2005}+\dfrac{x+6}{2004}+\dfrac{x+7}{2003}+3=0\)

\(\Leftrightarrow\dfrac{x+5}{2005}+1+\dfrac{x+6}{2004}+1+\dfrac{x+7}{2003}+1=0\)

\(\Leftrightarrow\dfrac{x+5+2005}{2005}+\dfrac{x+6+2004}{2004}+\dfrac{x+7+2003}{2003}=0\)

\(\Leftrightarrow\dfrac{x+2010}{2005}+\dfrac{x+2010}{2004}+\dfrac{x+2010}{2003}=0\)

\(\Leftrightarrow\left(x+2010\right)\left(\dfrac{1}{2005}+\dfrac{1}{2004}+\dfrac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2010=0\)

\(\Leftrightarrow x=-2010\)

Vậy ...