a) 1/4.x-1/3 = -5/9 b) 11/12-(2/5-x) = 3/4 c) 2007,5 -|x-1,5|= 0 d) (x-2)^2012 + |y^2 -9|^2014 = 0 e) 45-x/1963 + 40-x/1968 + 35-x/1973 + 30-x/1978

0 bình luận về “a) 1/4.x-1/3 = -5/9 b) 11/12-(2/5-x) = 3/4 c) 2007,5 -|x-1,5|= 0 d) (x-2)^2012 + |y^2 -9|^2014 = 0 e) 45-x/1963 + 40-x/1968 + 35-x/1973 + 30-x/1978”

1. Đáp án:

   $\begin{array}{l}
   a)\dfrac{1}{4}x – \dfrac{1}{3} =  – \dfrac{5}{9}\\
    \Rightarrow \dfrac{1}{4}x =  – \dfrac{5}{9} + \dfrac{1}{3}\\
    \Rightarrow \dfrac{1}{4}x =  – \dfrac{5}{9} + \dfrac{3}{9} =  – \dfrac{2}{9}\\
    \Rightarrow x = \dfrac{{ – 2}}{9}:\dfrac{1}{4}\\
    \Rightarrow x = \dfrac{{ – 8}}{9}\\
   b)\dfrac{{11}}{{12}} – \left( {\dfrac{2}{5} – x} \right) = \dfrac{3}{4}\\
    \Rightarrow \dfrac{{11}}{{12}} – \dfrac{2}{5} + x = \dfrac{3}{4}\\
    \Rightarrow x = \dfrac{3}{4} – \dfrac{{11}}{{12}} + \dfrac{2}{5}\\
    \Rightarrow x = \dfrac{{3.15 – 5.11 + 2.12}}{{60}}\\
    \Rightarrow x = \dfrac{{14}}{{60}} = \dfrac{7}{{30}}\\
   c)2007,5 – \left| {x – 1,5} \right| = 0\\
    \Rightarrow \left| {x – 1,5} \right| = 2007,5\\
    \Rightarrow \left[ \begin{array}{l}
   x – 1,5 = 2007,5\\
   x – 1,5 =  – 2007,5
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x = 2009\\
   x =  – 2006
   \end{array} \right.\\
   d){\left( {x – 2} \right)^{2012}} + {\left| {{y^2} – 9} \right|^{2014}} = 0\\
    \Rightarrow \left\{ \begin{array}{l}
   x – 2 = 0\\
   {y^2} – 9 = 0
   \end{array} \right.\\
    \Rightarrow \left\{ \begin{array}{l}
   x = 2\\
   {y^2} = 9
   \end{array} \right.\\
    \Rightarrow \left\{ \begin{array}{l}
   x = 2\\
   y = 3\\
   y =  – 3
   \end{array} \right.\\
   e)\dfrac{{45 – x}}{{1963}} + \dfrac{{40 – x}}{{1968}} + \dfrac{{35 – x}}{{1973}} + \dfrac{{30 – x}}{{1978}} + 4 = 0\\
    \Rightarrow \dfrac{{45 – x}}{{1963}} + 1 + \dfrac{{40 – x}}{{1968}} + 1\\
    + \dfrac{{35 – x}}{{1973}} + 1 + \dfrac{{30 – x}}{{1978}} + 1 = 0\\
    \Rightarrow \dfrac{{45 – x + 1963}}{{1963}} + \dfrac{{40 – x + 1968}}{{1968}}\\
    + \dfrac{{35 – x + 1973}}{{1973}} + \dfrac{{30 – x + 1978}}{{1978}} = 0\\
    \Rightarrow \dfrac{{2008 – x}}{{1963}} + \dfrac{{2008 – x}}{{1968}} + \dfrac{{2008 – x}}{{1973}} + \dfrac{{2008 – x}}{{1978}} = 0\\
    \Rightarrow \left( {2008 – x} \right).\left( {\dfrac{1}{{1963}} + \dfrac{1}{{1968}} + \dfrac{1}{{1973}} + \dfrac{1}{{1978}}} \right) = 0\\
    \Rightarrow x = 2008\\
   f)x – \dfrac{{20}}{{11.13}} – \dfrac{{20}}{{13.15}} – \dfrac{{20}}{{15.17}} – …\dfrac{{20}}{{53.55}} = \dfrac{3}{{11}}\\
    \Rightarrow x – 10.\left( {\dfrac{2}{{11.13}} + \dfrac{2}{{13.15}} + … + \dfrac{1}{{53.55}}} \right) = \dfrac{3}{{11}}\\
    \Rightarrow x – 10.\left( {\dfrac{1}{{11}} – \dfrac{1}{{55}}} \right) = \dfrac{3}{{11}}\\
    \Rightarrow x – 10.\dfrac{4}{{55}} = \dfrac{3}{{11}}\\
    \Rightarrow x = \dfrac{3}{{11}} + \dfrac{{40}}{{55}} = \dfrac{{55}}{{55}} = 1
   \end{array}$

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