b: ĐKXĐ: x>=1
\(PT\Leftrightarrow5\sqrt{x-1}-\dfrac{15}{2}\cdot\dfrac{\sqrt{x-1}}{3}=6+\sqrt{x-1}\)
=>\(\sqrt{x-1}\cdot\left(5-\dfrac{15}{2\cdot3}\right)-\sqrt{x-1}=6\)
=>\(\sqrt{x-1}\left(5-\dfrac{5}{2}-1\right)=6\)
=>\(\sqrt{x-1}\cdot\dfrac{3}{2}=6\)
=>\(\sqrt{x-1}=4\)
=>x-1=16
=>x=17(nhận)
c: ĐKXĐ: x>=5
\(PT\Leftrightarrow2\sqrt{x-5}-\dfrac{1}{3}\cdot3\sqrt{x-5}+\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
d: ĐKXĐ: x>=-1
\(PT\Leftrightarrow4\sqrt{x+1}-3\sqrt{x+1}+2\sqrt{x+1}=16-\sqrt{x+1}\)
=>\(3\sqrt{x+1}+\sqrt{x+1}=16\)
=>\(\sqrt{x+1}=4\)
=>x+1=16
=>x=15(nhận)