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\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\)\(\frac{x+1}{2003}\)
\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)\)\(=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\left(\frac{x+2004}{2000}\right)+\left(\frac{x+2004}{2001}\right)\)\(=\left(\frac{x+2004}{2002}\right)+\left(\frac{x+2004}{2003}\right)\)
\(\Leftrightarrow\left(x+2004\right)\)\(\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)\)\(=0\)
\(\Leftrightarrow x+2004=0\)
\(\Leftrightarrow x=-2004\)
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 : \(\frac{1}{2000}+\frac{1}{2001}>\frac{1}{2002}+\frac{1}{2003}\)
\(\Rightarrow x+2004=0\Rightarrow x=\left(-2004\right)\)
\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right).\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Mà \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
=> x + 2004 = 0
=> x = -2004
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
=> \(\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}\)
=> \(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
=> \(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
=> (x+2004).\(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
=> x+2004=0
=> x=-2004
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\Rightarrow x+2004=0\)
=>x=0-2004
=>x=-2004
vậy 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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(x+4/2000 + 1) + (x+3/2001 + 1) = (x+2/2002 + 1) + (x+1/2003 + 1)
x + 2004/2000 + x+2004/2001 = x+2004/2002 + x+2004/2003
(x + 2004) x (1/2000 + 1/2001) = (x + 2004) x (1/2002 + 1/2003)
Với x + 2004 = 0 => x = -2004, ta có: 0 x (1/2000 + 1/2001) = 0 x (1/2002 + 1/2003), đúng
Với x khác 0 => 1/2000 + 1/2001 = 1/2002 + 1/2003, vô lí
Vì 1/2000 + 1/2001 > 1/2002 + 1/2003
vậy x = -2004
\(\Leftrightarrow\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\).Do\(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(\Leftrightarrow x=-2004\)
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\left(\frac{x+4}{2000}+1\right)+\left(\frac{x+3}{2001}+1\right)=\left(\frac{x+2}{2002}+1\right)+\left(\frac{x+1}{2003}+1\right)\)
\(\left(\frac{x+4}{2000}+\frac{2000}{2000}\right)+\left(\frac{x+3}{2001}+\frac{2001}{2001}\right)=\left(\frac{x+2}{2002}+\frac{2002}{2002}\right)+\left(\frac{x+1}{2003}+\frac{2003}{2003}\right)\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(x+2004\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Ta thấy \(\frac{1}{2000}>\frac{1}{2001}>\frac{1}{2002}>\frac{1}{2003}\)
nên \(\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)\ne0\)
Do đó: x + 2004 = 0 => x = -2004
Vậy x = -2004
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Vì \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
Nên x + 2004 = 0
=> x = -2004
Vậy x = -2004
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}-\frac{x+2004}{2002}-\frac{x+2004}{2003}=0\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
Có: \(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\ne0\)
\(\Rightarrow x+2004=0\)
\(\Rightarrow x=-2004\)
\(\frac{x+4}{2000}+\frac{x+3}{2001}=\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Rightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+3}{2001}+1\)
\(\Rightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Rightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}-\frac{1}{2002}-\frac{1}{2003}\right)=0\)
có 1/2000 + 1/2001 - 1/2002 - 1/2003
=> x + 2004 = 0
=> x = -2004
\(\frac{x+4}{2000}+\frac{x+3}{2001}+\frac{x+2}{2002}+\frac{x+1}{2003}\)
\(\Leftrightarrow\frac{x+4}{2000}+1+\frac{x+3}{2001}+1=\frac{x+2}{2002}+1+\frac{x+1}{2003}+1\)
\(\Leftrightarrow\frac{x+2004}{2000}+\frac{x+2004}{2001}=\frac{x+2004}{2002}+\frac{x+2004}{2003}\)
\(\Leftrightarrow\left(x+2004\right)\left(\frac{1}{2000}+\frac{1}{2001}+\frac{1}{2002}+\frac{1}{2003}\right)=0\)
\(\Leftrightarrow x+2004=0\)
\(\Leftrightarrow x=-2004\)