e: ĐKXĐ: x>=2
Ta có: \(x-3\sqrt{x-2}-12=0\)
=>\(x-2-3\sqrt{x-2}-10=0\)
=>\(\left(\sqrt{x-2}-5\right)\left(\sqrt{x-2}+2\right)=0\)
=>\(\sqrt{x-2}-5=0\)
=>\(\sqrt{x-2}=5\)
=>x-2=25
=>x=27(nhận)
i: ĐKXĐ: x>=2
Ta có: \(\sqrt{36x-72}-15\cdot\sqrt{\frac{x-2}{25}}=4\left(5-\sqrt{x-2}\right)\)
=>\(6\sqrt{x-2}-15\cdot\frac{\sqrt{x-2}}{5}=20-4\sqrt{x-2}\)
=>\(3\sqrt{x-2}+4\sqrt{x-2}=20\)
=>\(7\cdot\sqrt{x-2}=20\)
=>\(\sqrt{x-2}=\frac{20}{7}\)
=>\(x-2=\frac{400}{49}\)
=>\(x=2+\frac{400}{49}=\frac{498}{49}\) (nhận)
k: ĐKXĐ: x>=5
Ta có: \(\sqrt{4x-20}+\sqrt{x-5}-\frac13\cdot\sqrt{9x-45}=4\)
=>\(2\sqrt{x-5}+\sqrt{x-5}-\frac13\cdot3\sqrt{x-5}=4\)
=>\(2\sqrt{x-5}=4\)
=>\(\sqrt{x-5}=2\)
=>x-5=4
=>x=9(nhận)
