a) x/7=3/x b) x+5/3 = 5/9 c) 1/x+ y/2= 5/8 d) /2x-1/+5= 8

0 bình luận về “a) x/7=3/x b) x+5/3 = 5/9 c) 1/x+ y/2= 5/8 d) /2x-1/+5= 8”

1. Đáp án:

   $\begin{array}{l}
   a)\dfrac{x}{7} = \dfrac{3}{x}\\
    \Rightarrow x.x = 3.7\\
    \Rightarrow {x^2} = 21\\
   Vậy\,{x^2} = 21\\
   b)\dfrac{{x + 5}}{3} = \dfrac{5}{9}\\
    \Rightarrow 9.\left( {x + 5} \right) = 3.5\\
    \Rightarrow x + 5 = \dfrac{5}{3}\\
    \Rightarrow x = \dfrac{5}{3} – 5\\
    \Rightarrow x = \dfrac{{ – 10}}{3}\\
   Vậy\,x = \dfrac{{ – 10}}{3}\\
   c)\dfrac{1}{x} + \dfrac{y}{2} = \dfrac{5}{8}\\
    \Rightarrow \dfrac{{2 + x.y}}{{2.x}} = \dfrac{5}{8}\\
    \Rightarrow 16 + 8.x.y = 10x\\
    \Rightarrow 8xy – 10x =  – 16\\
    \Rightarrow 2x.\left( {4y – 5} \right) =  – 8\\
    \Rightarrow x.\left( {4y – 5} \right) =  – 4\\
    \Rightarrow x.\left( {5 – 4y} \right) = 4 = 1.4 = 2.2
   \end{array}$

   $\begin{array}{l}
    \Rightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x = 4\\
   5 – 4y = 1
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x =  – 4\\
   5 – 4y =  – 1
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x = 1\\
   5 – 4y = 4
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x =  – 1\\
   5 – 4y =  – 4
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x = 2\\
   5 – 4y = 2
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x =  – 2\\
   5 – 4y =  – 2
   \end{array} \right.
   \end{array} \right. \Rightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x = 4\\
   4y = 4
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x =  – 4\\
   4y = 6
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x = 1\\
   4y = 1
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x =  – 1\\
   4y = 9
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x = 2\\
   4y = 3
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x =  – 2\\
   4y = 7
   \end{array} \right.
   \end{array} \right. \Rightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x = 4\\
   y = 1
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x =  – 4\\
   y = \dfrac{3}{2}
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x = 1\\
   y = \dfrac{1}{4}
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x =  – 1\\
   y = \dfrac{9}{4}
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x = 2\\
   y = \dfrac{3}{4}
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x =  – 2\\
   y = \dfrac{7}{4}
   \end{array} \right.
   \end{array} \right.\\
   Vậy\left( {x;y} \right) = \left\{ \begin{array}{l}
   \left( {4;1} \right);\left( { – 4;\dfrac{3}{2}} \right);\left( {1;\dfrac{1}{4}} \right);\\
   \left( { – 1;\dfrac{9}{4}} \right);\left( {2;\dfrac{3}{4}} \right);\left( { – 2;\dfrac{7}{4}} \right)
   \end{array} \right\}
   \end{array}$

   $\begin{array}{l}
   d)\left| {2x – 1} \right| + 5 = 8\\
    \Rightarrow \left| {2x – 1} \right| = 3\\
    \Rightarrow \left[ \begin{array}{l}
   2x – 1 = 3\\
   2x – 1 =  – 3
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   2x = 4\\
   2x =  – 2
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   x = 2\\
   x =  – 1
   \end{array} \right.\\
   Vậy\,x = 2;x =  – 1
   \end{array}$

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