A=\(\frac{x+1}{\sqrt{x}}-\left(\frac{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}-1\right)}-\frac{\left(\sqrt{x}+1\right)\left(x-\sqrt{x}+1\right)}{\sqrt{x}\left(\sqrt{x}+1\right)}\right)\)
  =\(\frac{x+1}{\sqrt{x}}-\left(\frac{x+\sqrt{x}+1}{\sqrt{x}}-\frac{x-\sqrt{x}+1}{\sqrt{x}}\right)\)
  =\(\frac{x+1}{\sqrt{x}}-\frac{x+\sqrt{x}+1-x+\sqrt{x}-1}{\sqrt{x}}\)
  =\(\frac{x+1}{\sqrt{x}}-\frac{2\sqrt{x}}{\sqrt{x}}\)
  =\(\frac{x+1-2\sqrt{x}}{\sqrt{x}}\)
  A =\(\frac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}}\)

\(\sqrt{7-2\sqrt{2+5\sqrt{2}+\sqrt{18-2\cdot4\cdot\sqrt{2}}}}\)=\(\sqrt{7-2\sqrt{2+5\sqrt{2}+4-\sqrt{2}}}\)

=\(\sqrt{7-2\sqrt{6+4\sqrt{2}}}=\sqrt{7-2\left(2+\sqrt{2}\right)}\) =\(\sqrt{3+2\sqrt{2}}\) =\(\sqrt{2}+1\)

\(\sqrt{10+2\sqrt{17-4\left(2+\sqrt{5}\right)}}\)=\(\sqrt{10+2\sqrt{17-8-4\sqrt{5}}}=\sqrt{10+2\sqrt{9-4\sqrt{5}}}\)

=\(\sqrt{10+2\left(\sqrt{5}-2\right)}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)

\(\sqrt{\sqrt{5-\sqrt{3-\sqrt{29-2\cdot3\cdot2\sqrt{5}}}}}=\)\(\sqrt{\sqrt{5-\sqrt{3-2\sqrt{5}+3}}}=\sqrt{\sqrt{5-\sqrt{6-2\sqrt{5}}}}\)=\(\sqrt{\sqrt{5-\sqrt{5}+1}}=\sqrt{1}=1\)

xin chào em mới học dưới lớp tám thôi khó quá không biết làm 

Ở đây em chỉ lp 7 thoy à

Nhưng mà em chỉ cho nhé

Qua hok 24 thì có thầy giải cho

tk nha