1: ĐKXĐ: x>=5
Ta có: \(\sqrt{x+4+6\sqrt{x-5}}+\sqrt{x+4\sqrt{x-5}-1}=\sqrt{9x-45}\)
=>\(\sqrt{x-5+2\cdot\sqrt{x-5}\cdot3+9}+\sqrt{x-5+2\cdot\sqrt{x-5}\cdot2+4}=3\sqrt{x-5}\)
=>\(\sqrt{\left(\sqrt{x-5}+3\right)^2}+\sqrt{\left(\sqrt{x-5}+2\right)^2}=3\sqrt{x-5}\)
=>\(\sqrt{x-5}+3+\sqrt{x-5}+2=3\sqrt{x-5}\)
=>\(\sqrt{x-5}=5\)
=>x-5=25
=>x=30(nhận)
2: ĐKXĐ: x>=3
Ta có: \(\sqrt{x+4\sqrt{x-3}+1}+\sqrt{x-4\sqrt{x-3}+1}=x-11\)
=>\(\sqrt{x-3+2\cdot\sqrt{x-3}\cdot2+4}+\sqrt{x-3-2\cdot\sqrt{x-3}\cdot2+4}=x-11\)
=>\(\sqrt{\left(\sqrt{x-3}+2\right)^2}+\sqrt{\left(\sqrt{x-3}-2\right)^2}=x-11\)
=>\(\left|\sqrt{x-3}+2\right|+\left|\sqrt{x-3}-2\right|=x-11\)
=>\(\sqrt{x-3}+2+\left|\sqrt{x-3}-2\right|=x-11\) (1)
TH1: x>=7
=>\(x-3\ge4\)
=>\(\sqrt{x-3}\ge2\)
=>\(\sqrt{x-3}-2\ge0\)
(1) sẽ trở thành: \(\sqrt{x-3}+2+\sqrt{x-3}-2=x-11\)
=>\(x-11=2\sqrt{x-3}\)
=>\(x-3-2\sqrt{x-3}-8=0\)
=>\(\left(\sqrt{x-3}-4\right)\left(\sqrt{x-3}+2\right)=0\)
=>\(\sqrt{x-3}-4=0\)
=>\(\sqrt{x-3}=4\)
=>x-3=16
=>x=19(nhận)
TH2: 3<=x<7
=>\(0\le\sqrt{x-3}<4\)
=>\(-2\le\sqrt{x-3}-2\le0\)
(1) sẽ trở thành: \(x-11=\sqrt{x-3}+2+2-\sqrt{x-3}=4\)
=>x=4+11=15(loại)
Vậy: x=19
