Thiếu không cậu?

x+4/2000 + x+3/2001 + x+2/2002 + x+1/2003

Sai đề ạ

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

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

<=> \(\frac{x-2004}{2000}+\frac{x-2004}{2001}+\frac{x-2004}{2002}=\frac{x-2004}{2}+\frac{x-2004}{3}+\frac{x-2004}{4}\)

<=> (x - 2004)(1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4) = 0

<=> x - 2004 = 0 (vì 1/2000 + 1/2001 + 1/2002 - 1/2 - 1/3 - 1/4 khác 0)

<=> x = 2004

Vậy S = {2004}

đề bài \(=\frac{x-2002}{2}+\frac{x-2001}{3}+\frac{x-2000}{4}\)

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

\(\Leftrightarrow\frac{x}{2000}-\frac{1}{500}+\frac{x}{2001}-\frac{1}{667}+\frac{x}{2002}-\frac{1}{1001}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}+1001+667+500=0\)

\(\Leftrightarrow\left(\frac{x}{2000}+\frac{x}{2001}+\frac{x}{2002}-\frac{x}{2}-\frac{x}{3}-\frac{x}{4}\right)+\left(1001+667+500-\frac{1}{500}-\frac{1}{667}-\frac{1}{1001}\right)=0\)

=> x=1

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

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

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

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

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

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

mà \(\dfrac{1}{2000}\) + \(\dfrac{1}{2001}\) - \(\dfrac{1}{2002}\) - \(\dfrac{1}{2003}\) khác 0

nên x+2004=0

=>x=0-2004
=> x = -2004
vậy S = -2004.

Tick nha

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

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

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

Với \(x-2004\ne0\)

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

Với \(x-2004=0\)

\(\Rightarrow x=2004\)

\(\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}=0\)

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

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

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

<=> x + 2004 = 0

<=> x = -2004

(Bn nhớ thêm kết quả là 0 vào sau nữa nha)

Mik sửa:

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

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

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

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

<=> x + 2004 = 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.

Các bạn vi phạm 3 điều trên sẽ bị giáo viên của Online Math trừ hết điểm hỏi đáp, có thể bị khóa tài khoản hoặc bị cấm vĩnh viễn không đăng nhập vào trang web.

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? NGUYỄN XUÂN TOÀN

(x+4/2000+1)+(x+3/2001+1)=(x+2/2002+1)+(x+1/2003+1)

=> (x+4/2000+2000/2000)+(x+3/2001+2001/2001)=(x+2/2002+2002/2002)+(x+1/2003+2003/2003)

=> x+2004/2000 + x+2004/2001= x+2004/2002 + x+2004/2003

=> (x+2004) . (1/2000 + 1/2001 - 1/2002 - 1/2003) = 0

=>                    x+2004 = 0

=>                              x= (-2004)

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

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

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

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

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

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

<=>   \(x=-2004\)