Bài 1 :
$\begin{array}{l}
a)\,\,{\left( {x - 2,5} \right)^4} = {\left( {x - 2,5} \right)^2}\\
\Rightarrow {\left( {x - 2,5} \right)^4} - {\left( {x - 2,5} \right)^2} = 0\\
\Rightarrow {\left( {x - 2,5} \right)^2}\left[ {{{\left( {x - 2,5} \right)}^2} - 1} \right] = 0\\
\Rightarrow \left[ \begin{array}{l}
{\left( {x - 2,5} \right)^2} = 0\\
{\left( {x - 2,5} \right)^2} - 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x - 2,5 = 0\\
x - 2,5 = 1
\end{array} \right. \Rightarrow \left[ \begin{array}{l}
x = 2,5\\
x = 3,5
\end{array} \right.\\
b)\,\,x\sqrt {x - 2} = 0\,\,\,\,\,\,\,\left( {DK:\,\,x \ge 2} \right)\\
\Rightarrow \left[ \begin{array}{l}
x = 0\,\,\,\,\,\left( {KTMDK} \right)\\
\sqrt {x - 2} = 0
\end{array} \right. \Rightarrow x - 2 = 0 \Rightarrow x = 2
\end{array}$
Bài 2 :
a) Ta có : $3 = \sqrt 9 $. Mà \(\sqrt 6 < \sqrt 9 \). Do đó \(\sqrt 6 < 3.\)
b) Ta có : \(\sqrt 7 < \sqrt 9 \)
\(\sqrt 15 < \sqrt {16} \)
\(\Rightarrow \sqrt 7 +\sqrt {15} <\sqrt {9}+\sqrt {16} = 3 + 4 = 7 \)
Vậy \(\sqrt 7 +\sqrt {15} <7.\)
