Question

\(Tìm\) \(x,\) \(biết\) \(:\)

\(a)\frac{x+1}{10}+\frac{x+1}{11}+\frac{x+1}{12}+=\frac{x+1}{13}+\frac{x+1}{14}\\ b)\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

Answer 1

( Câu trả lời bằng hình ảnh )

Answer 2

Tham khảo:

Answer 3

b) \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)

\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)

\(\Leftrightarrow\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)

\(\Leftrightarrow x+2004=0\)

\(\Leftrightarrow x=0-2004\)

\(\Rightarrow x=-2004\)

Vậy \(x=-2004.\)

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