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

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

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

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

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

<=>  \(x+2004=0\)  (do  1/2000 + 1/2001 - 1/2002 - 1/2003 khác 0)

<=>   \(x=-2004\)

I. Nội qui tham gia "Giúp tôi giải toán"

1. Không đưa câu hỏi linh tinh lên diễn đàn, chỉ đưa các bài mà mình không giải được hoặc các câu hỏi hay lên diễn đàn;

2. Không trả lời linh tinh, không phù hợp với nội dung câu hỏi trên diễn đàn.

3. Không "Đúng" vào các câu trả lời linh tinh nhằm gian lận điểm hỏi đáp.

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tôi mong các bn đừng làm như vậy !!!

nói như cặc mà cũng nói được

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

\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)\)\(=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2004}{2000}\right)+\left(\frac{x+2004}{2001}\right)\)\(=\left(\frac{x+2004}{2002}\right)+\left(\frac{x+2004}{2003}\right)\)
\(\Leftrightarrow\left(x+2004\right)\)\(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)\(=0\)

\(\Leftrightarrow x+2004=0\)
\(\Leftrightarrow x=-2004\)

Đúng ko vậy

Do vo van hoa

Sửa lại đề : Tìm x biết : \(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)

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

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

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

\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2014}{2002}-\frac{x+2004}{2003}=0\)

\(\Rightarrow\left(x+2004\right)\left[\left(\frac{1}{2000}+\frac{1}{2001}\right)-\left(\frac{1}{2002}+\frac{1}{2003}\right)\right]=0\)

Mà : \(\frac{1}{2000}+\frac{1}{2001}>\frac{1}{2002}+\frac{1}{2003}\)

\(\Rightarrow x+2004=0\Rightarrow x=\left(-2004\right)\)

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

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

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

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

\(\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)\)

⇔\(x+2014=0\)

⇔\(x=-2014\)

\(\Rightarrow\left(\dfrac{x+4}{2000}+1\right)+\left(\dfrac{x+3}{2001}+1\right)=\left(\dfrac{x+2}{2002}+1\right)+\left(\dfrac{x+1}{2003}+1\right)\\ \Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}-\dfrac{x+2004}{2002}-\dfrac{x+2004}{2003}=0\\ \Rightarrow\left(x+2004\right)\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\right)=0\\ \Rightarrow x=-2004\left(\dfrac{1}{2000}+\dfrac{1}{2001}-\dfrac{1}{2002}-\dfrac{1}{2003}\ne0\right)\)

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

\(\Leftrightarrow x+2004=0\)

hay x=-2004

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

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

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

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

có 1/2000 + 1/2001 - 1/2002 - 1/2003

=> x + 2004 = 0

=> x = -2004

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

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

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

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

\(\Leftrightarrow x+2004=0\)

\(\Leftrightarrow x=-2004\)

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

<=> \(\frac{x+4}{2000}+\frac{x+3}{2001}-\frac{x+2}{2002}-\frac{x+1}{2003}=0\)

<=> \(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)-\left(\frac{x+2}{2002}+1\right)-\left(\frac{x+1}{2003}+1\right)=0\)

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

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

<=> x+2004 =0 ( do \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)

<=> x= -2004

Học tốt

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

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

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

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

Vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}+\frac{1}{2003}\ne0\)

=> x + 2004 =0

=> x             = -2004