ĐK: x≥-1
Ta có: \(\sqrt{2x+3}+\sqrt{2x+2}=1\)
\(\Leftrightarrow4x+5+2\sqrt{\left(2x+3\right)\left(2x+2\right)}=1\)
\(\Leftrightarrow\sqrt{4x^2+10x+6}=-2x-2\) ĐK:x≤1
\(\Leftrightarrow4x^2+10x+6=4x^2+8x+4\)
\(\Leftrightarrow2x=-2\Leftrightarrow x=-1\left(tm\right)\)
\(\sqrt{2x+3}+\sqrt{2x+2}=1\)
<=> \(\sqrt{2x+3}=1-\sqrt{2x+2}\)
<=> 2x + 3 = (1 - \(\sqrt{2x+2}\))2
<=> 2x + 3 = 1 - 2\(\sqrt{2x+2}\) + (2x + 2)
<=> 2\(\sqrt{2x+2}\) = 2x - 2x + 1 + 2 - 3
<=> \(2\sqrt{2x+2}\) = 0
<=> \(\sqrt{2x+2}\) = 0
<=> 2x + 2 = 0
<=> 2x = -2
<=> x = -1