a) $\sqrt{x^2 +1}$ + $\sqrt{4x^2 -4x+5}$ =0 b) $\sqrt{2x^2 -4x +3}$ + $\sqrt{3x^2 -6x+7}$ = 2-x^2 +2x c) $\sqrt{x-1}$ / x + x/$\sqrt{x -1}$ =2

0 bình luận về “a) $\sqrt{x^2 +1}$ + $\sqrt{4x^2 -4x+5}$ =0 b) $\sqrt{2x^2 -4x +3}$ + $\sqrt{3x^2 -6x+7}$ = 2-x^2 +2x c) $\sqrt{x-1}$ / x + x/$\sqrt{x -1}$ =2”

1. \(\begin{array}{l}
   a)\,\sqrt {{x^2} + 1} + \sqrt {{{\left( {2x – 1} \right)}^2} + 4} = 0\\
   Vi\,\,\left\{ \begin{array}{l}
   \sqrt {{x^2} + 1} \ge 1\\
   \sqrt {{{\left( {2x – 1} \right)}^2} + 4} \ge 4
   \end{array} \right.\\
   \Rightarrow \sqrt {{x^2} + 1} + \sqrt {{{\left( {2x – 1} \right)}^2} + 4} \ge 5 > 0\\
   \Rightarrow PTVN
   \end{array}\)
   \(\begin{array}{l}
   b)\,pt \Leftrightarrow \sqrt {2{{\left( {x – 1} \right)}^2} + 1} + \sqrt {3{{\left( {x – 1} \right)}^2} + 4} = 3 – {\left( {x – 1} \right)^2}\left( * \right)\\
   Ta\,co:\,\,\sqrt {2{{\left( {x – 1} \right)}^2} + 1} + \sqrt {3{{\left( {x – 1} \right)}^2} + 4} \ge 1 + 2 = 3\\
   3 – {\left( {x – 1} \right)^2} \le 3\\
   \Rightarrow \left( * \right)\,xay\,ra\,khi\,x – 1 = 0 \Leftrightarrow x = 1
   \end{array}\)

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