Câu hỏi của Sách Giáo Khoa

Question

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

Sáng · Accepted Answer

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

$\Leftrightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}$

$\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$

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

$\Leftrightarrow x+2004=0$

$\Leftrightarrow x=-2004$Câu trả lời: $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

$\Leftrightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}$

$\Leftrightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$

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

$\Leftrightarrow x+2004=0$

$\Leftrightarrow x=-2004$

Trần Minh Hoàng · Answer

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

$\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1$

$\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$

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

Mà $\dfrac{1}{2000}+\dfrac{1}{2001}\ne\dfrac{1}{2002}+\dfrac{1}{2003}$ nên $x+2004=0\Rightarrow x=-2004$

Vậy, x = -2004Câu trả lời: $\dfrac{x+4}{2000}+\dfrac{x+3}{2001}=\dfrac{x+2}{2002}+\dfrac{x+1}{2003}$

$\Rightarrow\dfrac{x+4}{2000}+1+\dfrac{x+3}{2001}+1=\dfrac{x+2}{2002}+1+\dfrac{x+1}{2003}+1$

$\Rightarrow\dfrac{x+2004}{2000}+\dfrac{x+2004}{2001}=\dfrac{x+2004}{2002}+\dfrac{x+2004}{2003}$

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

Mà $\dfrac{1}{2000}+\dfrac{1}{2001}\ne\dfrac{1}{2002}+\dfrac{1}{2003}$ nên $x+2004=0\Rightarrow x=-2004$

Vậy, x = -2004