b:ĐKXĐ: x>=1\(4\sqrt{x+3}-\sqrt{x-1}=x+7\)=>\(4\sqrt{x+3}-8+\sqrt{x-1}=x-1\)=>\(4\left(\sqrt{x+3}-2\right)+\sqrt{x-1}-\left(x-1\right)=0\)=>\(4\cdot\frac{x+3-4}{\sqrt{x+3}+2}+\sqrt{x-1}-\left(x-1\right)=0\)=>\(\sqrt{x-1}\left(\frac{4\sqrt{x-1}}{\sqrt{x+3}+2}+1-\sqrt{x-1}\right)=0\)=>\(\sqrt{x-1}=0\)=>x-1=0=>x=1(nhận)c: ĐKXĐ: 4 \(\sqrt{x-4}-1+\sqrt{6-x}=1=x^2-10x+25\)=>\(\frac{x-4-1}{\sqrt{x-4}+1}+\frac{6-x-1}{\sqrt{6-x}+1}=\left(x-5\right)^2\)=>\(\frac{x-5}{\sqrt{x-4}+1}-\frac{x-5}{\sqrt{6-x}+1}-\left(x-5\right)^2=0\)=>\(\left(x-5\right)\left(\frac{1}{\sqrt{x-4}+1}-\frac{1}{\sqrt{6-x}+1}-x+5\right)=0\)=>x-5=0=>x=5(nhận)d: ĐKXĐ: -2 \(\sqrt{x+2}-2+\sqrt{6-x}-2=x^2-4x+4\)=>\(\frac{x+2-4}{\sqrt{x+2}+2}+\frac{6-x-4}{\sqrt{6-x}+2}=\left(x-2\right)^2\)=>\(\left(x-2\right)\cdot\left(\frac{1}{\sqrt{x+2}+2}-\frac{1}{\sqrt{6-x}+2}-x+2\right)=0\)=>x-2=0=>x=2(nhận)Câu trả lời: b:ĐKXĐ: x>=1\(4\sqrt{x+3}-\sqrt{x-1}=x+7\)=>\(4\sqrt{x+3}-8+\sqrt{x-1}=x-1\)=>\(4\left(\sqrt{x+3}-2\right)+\sqrt{x-1}-\left(x-1\right)=0\)=>\(4\cdot\frac{x+3-4}{\sqrt{x+3}+2}+\sqrt{x-1}-\left(x-1\right)=0\)=>\(\sqrt{x-1}\left(\frac{4\sqrt{x-1}}{\sqrt{x+3}+2}+1-\sqrt{x-1}\right)=0\)=>\(\sqrt{x-1}=0\)=>x-1=0=>x=1(nhận)c: ĐKXĐ: 4 \(\sqrt{x-4}-1+\sqrt{6-x}=1=x^2-10x+25\)=>\(\frac{x-4-1}{\sqrt{x-4}+1}+\frac{6-x-1}{\sqrt{6-x}+1}=\left(x-5\right)^2\)=>\(\frac{x-5}{\sqrt{x-4}+1}-\frac{x-5}{\sqrt{6-x}+1}-\left(x-5\right)^2=0\)=>\(\left(x-5\right)\left(\frac{1}{\sqrt{x-4}+1}-\frac{1}{\sqrt{6-x}+1}-x+5\right)=0\)=>x-5=0=>x=5(nhận)d: ĐKXĐ: -2 \(\sqrt{x+2}-2+\sqrt{6-x}-2=x^2-4x+4\)=>\(\frac{x+2-4}{\sqrt{x+2}+2}+\frac{6-x-4}{\sqrt{6-x}+2}=\left(x-2\right)^2\)=>\(\left(x-2\right)\cdot\left(\frac{1}{\sqrt{x+2}+2}-\frac{1}{\sqrt{6-x}+2}-x+2\right)=0\)=>x-2=0=>x=2(nhận)