Question

Giải các phương trình sau :

a) \(\dfrac{201-x}{99}\) + \(\dfrac{203-x}{97}\) = \(\dfrac{205-x}{95}\) + 3 = 0

b) \(\dfrac{2-x}{2002}\) - 1 = \(\dfrac{1-x}{2003}\) - \(\dfrac{x}{2004}\)

Answer 1

a)\(\dfrac{201-x}{99}+\dfrac{203-x}{97}=\dfrac{205-x}{95}+3=0\)

<=>\(\left(\dfrac{201-x}{99}+1\right)+\left(\dfrac{203-x}{97}+1\right)+\left(\dfrac{205-x}{95}+1\right)=0\)

<=>\(\dfrac{201-x+99}{99}+\dfrac{203-x+97}{97}=\dfrac{205-x+95}{95}=0\)

<=> \(\dfrac{300-x}{99}+\dfrac{300-x}{97}=\dfrac{300-x}{95}=0\)

<=> \(\left(300-x\right)\left(\dfrac{1}{99}+\dfrac{1}{97}+\dfrac{1}{95}\right)=0\)

<=> 300 - x = 0

<=> x = 300

b) \(\dfrac{2-x}{2002}-1=\dfrac{1-x}{2003}-\dfrac{x}{2004}\)

<=> \(\dfrac{2-x}{2002}+1=\left(\dfrac{1-x}{2003}+1\right)+\left(\dfrac{x}{2004}+1\right)\){Cộng cả hai vế của phương trình với 2}

<=> \(\dfrac{2-x+2002}{2002}=\dfrac{1-x+2003}{2003}+\dfrac{-x+2004}{2004}\)

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

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

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

<=> 2004 - x = 0

<=> x = 2004.

Answer 2

b) \(\dfrac{2-x}{2002}+\dfrac{x}{2004}-1=\dfrac{1-x}{2003}\)

\(\Leftrightarrow\dfrac{2-x}{2002}+1+\dfrac{x}{2004}-1=\dfrac{1-x}{2003}+1\)( cộng 2 vế cho 1)

\(\Leftrightarrow\dfrac{2-x+2002}{2002}+\dfrac{x-2004}{2004}=\dfrac{1-x+2003}{2003}\)

\(\Leftrightarrow\dfrac{2004-x}{2002}+\dfrac{x-2004}{2004}=\dfrac{2004-x}{2003}\)

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

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

\(\Rightarrow x=2004\) do \(\left(\dfrac{-1}{2002}+\dfrac{1}{2004}+\dfrac{1}{2003}\ne0\right)\)