ĐKXĐ: \(x\ge\dfrac{3}{2}\)
\(\sqrt{2x-3+2\sqrt{2x-3}+1}+\sqrt{2x-3+8\sqrt{2x-3}+16}=7\)
\(\Leftrightarrow\sqrt{\left(\sqrt{2x-3}+1\right)^2}+\sqrt{\left(\sqrt{2x-3}+4\right)^2}=7\)
\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=7\)
\(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=7\)
\(\Leftrightarrow2\sqrt{2x-3}=2\)
\(\Leftrightarrow\sqrt{2x-3}=1\)
\(\Leftrightarrow x=2\)
ĐK: \(x\ge\dfrac{3}{2}\)
Ta có: \(\sqrt{2x-2+2\sqrt{2x-3}}+\sqrt{2x+13+8\sqrt{2x-3}}=7\)
\(\Leftrightarrow\left|\sqrt{2x-3}+1\right|+\left|\sqrt{2x-3}+4\right|=7\)
Vì \(\sqrt{2x-3}\ge0\) \(\Leftrightarrow\sqrt{2x-3}+1+\sqrt{2x-3}+4=7\)
\(\Leftrightarrow2\sqrt{2x-3}=2\)
\(\Leftrightarrow\sqrt{2x-3}=1\)
\(\Leftrightarrow2x-3=1\Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\)
