a: \(A=\frac{1}{\sqrt2+1}-\frac{\sqrt8-\sqrt{10}}{2-\sqrt5}\)
\(=\frac{1\left(\sqrt2-1\right)}{\left(\sqrt2+1\right)\left(\sqrt2-1\right)}-\frac{\sqrt2\left(2-\sqrt5\right)}{2-\sqrt5}\)
\(=\sqrt2-1-\sqrt2=-1\)
b:Sửa đề: \(A=\frac{2+\sqrt3}{\sqrt{7-4\sqrt3}}-\frac{2-\sqrt3}{\sqrt{7+4\sqrt3}}\)
\(=\frac{2+\sqrt3}{\sqrt{\left(2-\sqrt3\right)^2}}-\frac{2-\sqrt3}{\sqrt{\left(2+\sqrt3\right)^2}}\)
\(=\frac{2+\sqrt3}{2-\sqrt3}-\frac{2-\sqrt3}{2+\sqrt3}=\frac{\left(2+\sqrt3\right)^2-\left(2-\sqrt3\right)^2}{\left(2-\sqrt3\right)\left(2+\sqrt3\right)}\)
\(=7+4\sqrt3-\left(7-4\sqrt3\right)=8\sqrt3\)
c: \(P=\frac{\sqrt5}{\sqrt5-2}-2\sqrt5\)
\(=\frac{\sqrt5\left(\sqrt5+2\right)}{\left(\sqrt5+2\right)\left(\sqrt5-2\right)}-2\sqrt5\)
\(=\sqrt5\left(\sqrt5+2\right)-2\sqrt5=5+2\sqrt5-2\sqrt5=5\)
d: \(A=\frac{1}{3+\sqrt7}+\frac{1}{3-\sqrt7}\)
\(=\frac{3-\sqrt7+3+\sqrt7}{\left(3-\sqrt7\right)\left(3+\sqrt7\right)}\)
\(=\frac{6}{9-7}=\frac62=3\)
