`10)sqrt{5+sqrt{21+sqrt{80}}}=sqrt{5+sqrt{21+2.sqrt{20}+1}}=sqrt{5+sqrt{(sqrt{20}+1)^2}}=sqrt{5+1+sqrt{20}}=sqrt{6+sqrt{20}}=sqrt{5+2sqrt5+1}=sqrt{(sqrt5+1)^2}=sqrt5+1` $\\$`11)sqrt{2-sqrt3}+sqrt{2+sqrt3}=sqrt{(4-2sqrt3)/2}+sqrt{(4+2sqrt3)/2}=sqrt{(3-2sqrt3+1)/2}+sqrt{(3+2sqrt3+1)/2}=sqrt{(sqrt3-1)^2/2}+sqrt{(sqrt3+1)^2/2}=(sqrt3-1)/sqrt2+(sqrt3+1)/sqrt2=(2sqrt3)/sqrt2=sqrt6`(Rất đơn giản chỉ cần nhân 2 rồi chia 2 ở trong căn.) $\\$ 12 sai đề `3-5=-2<0` nên không tồn tại căn thức nên mình sẽ chuyển thành như sau:`12)sqrt{3-sqrt5}-sqrt{3+sqrt5}=sqrt{(6-2sqrt5)/2}-sqrt{(6+2sqrt5)/2}=sqrt{(5-2sqrt5+1)/2}+sqrt{(5+2sqrt5+1)/2}=sqrt{(sqrt5-1)^2/2}+sqrt{(sqrt5+1)^2/2}=(sqrt5-1)/sqrt2-(sqrt5+1)/sqrt2=(-2)/sqrt2=-sqrt2` $\\$ `13)sqrt{45+sqrt{2009}}+sqrt{45-sqrt{2009}}=sqrt{(90+2sqrt{2009})/2}+sqrt{(90-2sqrt{2009})/2}=sqrt{(49+2sqrt{41.49}+49)/2}+sqrt{(49-2sqrt{49.41}+41)/2}=sqrt{(7+sqrt{41})^2/2}+sqrt{(7-sqrt{41})^2/2}=(7+sqrt{41})/sqrt2+(7-sqrt{41})/2=(14)/sqrt2=7sqrt2` $\\$ `14)sqrt{4+sqrt7}-sqrt{4-sqrt7}-sqrt2=sqrt{(8+2sqrt7)/2}-sqrt{(8-2sqrt7)/2}-sqrt2=sqrt{(7+2sqrt7+1)/2}-sqrt{(7-2sqrt7+1)/2}-sqrt2=sqrt{(sqrt7+1)^2/2}-sqrt{(sqrt7-1)^2/2}-sqrt2=(sqrt7+1)/2-(sqrt7-1)/2-sqrt2=2/sqrt2-sqrt2=sqrt2-sqrt2=0`