Given a,b satisfying 2^a-6 . (7/2)^b= 28^a. Find the sum a +b

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1. $\text{Ta có:}$

   $2^{a – 6} . (\dfrac{7}{2})^{b} = 28^{a}$ 

   $⇒ 2^{a – 6} . \dfrac{7^{b}}{2^{b}} = (7 . 2^{2})^{a}$ 

   $⇒ 2^{a – 6 – a} . 7^{b} = 7^{a} . 2^{2a}$

   $⇒ a – 6 – b = 2a ; b = a$

   $⇒ a – 6 – a = 2a ; b = a$

   $⇒ -2a = 6 ; b = a$

   $⇒ a = -3 ; b = a = -3$

   $⇒ a + b = -3 + (-3) = -6$

   $\text{Vậy}$ $a + b = -6$

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2. Đáp án:

   $\begin{array}{l}
   {2^{a – 6}}.{\left( {\dfrac{7}{2}} \right)^b} = {28^a}\\
    \Rightarrow {2^{a – 6}}.\dfrac{{{7^b}}}{{{2^b}}} = {\left( {{2^2}.7} \right)^a}\\
    \Rightarrow {2^{a – 6 – b}}{.7^b} = {2^{2a}}{.7^a}\\
    \Rightarrow \left\{ \begin{array}{l}
   a – 6 – b = 2a\\
   b = a
   \end{array} \right.\\
    \Rightarrow \left\{ \begin{array}{l}
   a – 6 – a = 2a\\
   b = a
   \end{array} \right.\\
    \Rightarrow \left\{ \begin{array}{l}
   2a =  – 6\\
   b = a
   \end{array} \right.\\
    \Rightarrow \left\{ \begin{array}{l}
   a =  – 3\\
   b = a =  – 3
   \end{array} \right.\\
    \Rightarrow a + b =  – 3 + \left( { – 3} \right) =  – 6
   \end{array}$

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