a/ ĐKXĐ: \(x\ge 0,x\ne 1\)
\(P=\dfrac{\sqrt x+2}{\sqrt x-1}-\dfrac{6}{\sqrt x+1}+\dfrac{x-7}{x-1}\\=\dfrac{(\sqrt x+2)(\sqrt x+1)}{(\sqrt x-1)(\sqrt x+1)}-\dfrac{6(\sqrt x-1)}{(\sqrt x-1)(\sqrt x+1)}+\dfrac{x-7}{(\sqrt x-1)(\sqrt x+1)}\\=\dfrac{x+3\sqrt x+2-6\sqrt x+6+x-7}{(\sqrt x-1)(\sqrt x+1)}\\=\dfrac{2x-3\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}\\=\dfrac{2x-2\sqrt x-\sqrt x+1}{(\sqrt x-1)(\sqrt x+1)}\\=\dfrac{2\sqrt x(\sqrt x-1)-(\sqrt x-1)}{(\sqrt x-1)(\sqrt x+1)}\\=\dfrac{(2\sqrt x-1)(\sqrt x-1)}{(\sqrt x-1)(\sqrt x+1)}\\=\dfrac{2\sqrt x-1}{\sqrt x+1}\)
b/ \(\sqrt{10-2\sqrt{21}}\\=\sqrt{7-2\sqrt{21}+3}\\=\sqrt{(\sqrt 7-\sqrt 3)^2}\\=|\sqrt 7-\sqrt 3|\\=\sqrt 7-\sqrt 3\\→x=(\sqrt 7+\sqrt 3)(\sqrt 7-\sqrt 3)=7-3=4\)
Thay \(x=4(TM)\) vào biểu thức \(P\):
\(P=\dfrac{2.\sqrt 4-1}{\sqrt 4+1}=\dfrac{4-1}{2+1}=\dfrac{3}{3}=1\)
Vậy \(P=1\) tại \(x=4\)
