Bài 25 trang 9 SBT toán 8 tập 2

Đề bài

Giải các phương trình sau:

a) \(\dfrac{{2x}}{3} + \dfrac{{2x - 1}}{6} = 4 - \dfrac{x}{3}\)

b) \(\dfrac{{x - 1}}{2} + \dfrac{{x - 1}}{4} = 1 - \dfrac{{2\left( {x - 1} \right)}}{3}\)

c) \(\dfrac{{2 - x}}{{2001}} - 1 = \dfrac{{1 - x}}{{2002}} - \dfrac{x}{{2003}}\)

a)

\(\eqalign{
& {{2x} \over 3} + {{2x - 1} \over 6} = 4 - {x \over 3} \cr 
& \Leftrightarrow {{2.2x} \over 6} + {{2x - 1} \over 6} = {{4.6} \over 6} - {{2x} \over 6} \cr 
& \Leftrightarrow 2.2x + 2x - 1 = 4.6 - 2x \cr 
& \Leftrightarrow 4x + 2x - 1 = 24 - 2x \cr 
& \Leftrightarrow 6x - 1 = 24 - 2x \cr 
& \Leftrightarrow 6x + 2x = 24 + 1 \cr 
& \Leftrightarrow 8x = 25 \cr 
& \Leftrightarrow x = {{25} \over 8} \cr} \)

Vậy phương trình có nghiệm \(x = \dfrac{25} { 8}.\)

b)

\(\eqalign{
& {{x - 1} \over 2} + {{x - 1} \over 4} = 1 - {{2\left( {x - 1} \right)} \over 3} \cr 
& \Leftrightarrow {{x - 1} \over 2} + {{x - 1} \over 4} = 1 - {{2x - 2} \over 3} \cr} \)
\(\displaystyle \Leftrightarrow {{6\left( {x - 1} \right)} \over {12}} + {{3\left( {x - 1} \right)} \over {12}} \) \(\displaystyle = {{12} \over {12}} - {{4\left( {2x - 2} \right)} \over {12}} \)

\( \Leftrightarrow 6\left( {x - 1} \right) + 3\left( {x - 1} \right) \) \(= 12 - 4\left( {2x - 2} \right) \)
\( \Leftrightarrow 6x - 6 + 3x - 3 = 12 - 8x + 8 \)
\( \Leftrightarrow 6x + 3x + 8x = 12 + 8 + 6 + 3 \) 
\( \Leftrightarrow 17x = 29 \)
\(\displaystyle\Leftrightarrow x = {{29} \over {17}}\)

Vậy phương trình có nghiệm \(x = \dfrac{{29}}{{17}}.\)

c) 

\(\displaystyle {{2 - x} \over {2001}} - 1 = {{1 - x} \over {2002}} - {x \over {2003}} \)

\(\displaystyle \Leftrightarrow {{2 - x} \over {2001}} - 1 + 2 \) \(\displaystyle= {{1 - x} \over {2002}} + 1 + 1 - {x \over {2003}} \)
\(\displaystyle\Leftrightarrow {{2 - x} \over {2001}} + 1 \) \(\displaystyle= \left( {{{1 - x} \over {2002}} + 1} \right) + \left( {1 - {x \over {2003}}} \right) \)
\(\displaystyle \Leftrightarrow {{2003 - x} \over {2001}} \) \(\displaystyle= {{2003 - x} \over {2002}} + {{2003 - x} \over {2003}} \)
\(\displaystyle \Leftrightarrow {{2003 - x} \over {2001}} - {{2003 - x} \over {2002}} \) \(\displaystyle- {{2003 - x} \over {2003}} = 0 \) 
\(\displaystyle  \Leftrightarrow \left( {2003 - x} \right) \) \(\displaystyle\left( {{1 \over {2001}} - {1 \over {2002}} - {1 \over {2003}}} \right) = 0 \) 
\( \Leftrightarrow 2003 - x = 0 \)
\( \Leftrightarrow x = 2003 \)

(Vì \(\dfrac{1}{{2001}} - \dfrac{1}{{2002}} - \dfrac{1}{{2003}} \ne 0\).)

Vậy phương trình có nghiệm \(x = 2003.\)