Giải pt sqrt(x+2)+sqrt(5-x)+sqrt(x+2)(5-x)=4

0 bình luận về “Giải pt sqrt(x+2)+sqrt(5-x)+sqrt(x+2)(5-x)=4”

1. Đáp án: $x = \frac{{3 + 3\sqrt 5 }}{2}$

    

   Giải thích các bước giải:

   $\begin{array}{l}
   ĐKxđ: – 2 \le x \le 5\\
   \sqrt {x + 2}  + \sqrt {5 – x}  = a\left( {a > 0} \right)\\
    \Rightarrow {a^2} = x + 2 + 5 – x + 2\sqrt {x + 2} .\sqrt {5 – x} \\
    \Rightarrow {a^2} = 7 + 2\sqrt {\left( {x – 2} \right)\left( {5 – x} \right)} \\
    \Rightarrow \sqrt {\left( {x – 2} \right)\left( {5 – x} \right)}  = \frac{{{a^2} – 7}}{2}\\
    \Rightarrow pt:a + \frac{{{a^2} – 7}}{2} = 4\\
    \Rightarrow {a^2} + 2a – 7 – 8 = 0\\
    \Rightarrow {a^2} + 2a – 15 = 0\\
    \Rightarrow \left( {a – 3} \right)\left( {a + 5} \right) = 0\\
    \Rightarrow a = 3\left( {do\,:a > 0} \right)\\
    \Rightarrow \left\{ \begin{array}{l}
   \sqrt {x + 2}  + \sqrt {5 – x}  = 3\\
   \sqrt {\left( {x – 2} \right)\left( {5 – x} \right)}  = 1
   \end{array} \right. \Rightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   \sqrt {x + 2}  = \frac{{3 – \sqrt 5 }}{2}\\
   \sqrt {5 – x}  = \frac{{3 + \sqrt 5 }}{2}
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   \sqrt {x + 2}  = \frac{{3 + \sqrt 5 }}{2}\\
   \sqrt {5 – x}  = \frac{{3 – \sqrt 5 }}{2}
   \end{array} \right.
   \end{array} \right.\\
    \Rightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x + 2 = \frac{{7 – 3\sqrt 5 }}{2}\\
   5 – x = \frac{{7 + 3\sqrt 5 }}{2}
   \end{array} \right.\\
   \left\{ \begin{array}{l}
   x + 2 = \frac{{7 + 3\sqrt 5 }}{2}\\
   5 – x = \frac{{7 – 3\sqrt 5 }}{2}
   \end{array} \right.
   \end{array} \right. \Rightarrow \left[ \begin{array}{l}
   \left\{ \begin{array}{l}
   x = \frac{{1 – 3\sqrt 5 }}{2}\\
   x = \frac{{3 – 3\sqrt 5 }}{2}
   \end{array} \right.\left( {ktm} \right)\\
   \left\{ \begin{array}{l}
   x = \frac{{3 + 3\sqrt 5 }}{2}\\
   x = \frac{{3 + 3\sqrt 5 }}{2}
   \end{array} \right.
   \end{array} \right. \Rightarrow x = \frac{{3 + 3\sqrt 5 }}{2}\left( {tm} \right)
   \end{array}$

   Bình luận