Giải thích các bước giải:
a.$\sqrt{y^6}=\sqrt{(y^3)^2}=|y^3|$
b.$\sqrt{64x^8}=\sqrt{(8x^4)^2}=8x^4$
c.$x-\sqrt{1-2x+x^2}=x-\sqrt{(x-1)^2}=x-|x-1|=x-(x-1)=1$ vì $x>1\to x-1>0$
d.$\sqrt{x-6\sqrt{x-9}}=\sqrt{x-9-6\sqrt{x-9}+9}=\sqrt{(\sqrt{x-9}-3)^2}=|\sqrt{x-9}-3|$
Vì $9\le x\le 18$
$\to 0\le x-9\le 9$
$\to 0\le\sqrt{x-9}\le 3$
$\to \sqrt{x-9}-3\le 0$
$\to |\sqrt{x-9}-3|=-\sqrt{x-9}+3$
$\to \sqrt{x-6\sqrt{x-9}}=-\sqrt{x-9}+3$