a)
\(\eqalign{
& \sqrt {2x + 3} = 1 + \sqrt 2\cr & \Leftrightarrow 2x + 3 = {(1 + \sqrt 2 )^2} \cr
& \Leftrightarrow 2x + 3 = 1 + 2\sqrt 2 + 2 \cr} \)
\(\Leftrightarrow 2x=2\sqrt 2 \Leftrightarrow x= \sqrt 2\)
b) \(\sqrt {10 + \sqrt {3}x } = 2 + \sqrt 6 \)
\( \Leftrightarrow 10 + \sqrt {3}x = {(2 + \sqrt 6 )^2}\)
\( \Leftrightarrow 10 + \sqrt {3}x = 4 + 4\sqrt 6 + 6\)\( \Leftrightarrow \sqrt {3}x = 4\sqrt 6 \)
\( \displaystyle \Leftrightarrow x = {{4\sqrt 6 } \over {\sqrt 3 }} \Leftrightarrow x = 4\sqrt 2 \)
c)
\(\eqalign{
& \sqrt {3x - 2} = 2 - \sqrt 3\cr & \Leftrightarrow 3x - 2 = {(2 - \sqrt 3 )^2} \cr
& \Leftrightarrow 3x - 2 = 4 - 4\sqrt 3 + 3 \cr} \)
\( \displaystyle\Leftrightarrow 3x = 9 - 4\sqrt 3 \Leftrightarrow x = {{9 - 4\sqrt 3 } \over 3}\)
d) \(\sqrt {x + 1} = \sqrt 5 - 3\)
Ta có:
\(\sqrt 5 <\sqrt 9 \) \( \Leftrightarrow \sqrt 5 < 3 \Leftrightarrow \sqrt 5 - 3 < 0\)
Không có giá trị nào của \(x\) để \(\sqrt {x + 1} = \sqrt 5 - 3\).
