Đáp án:
$7)\dfrac{3n}{2} \\ 8)-\dfrac{\sqrt{2}}{4a}\\ 9)-\sqrt{3}\\ 10)\dfrac{3(a-3)}{4} \\ 11)-\dfrac{2a+3}{b}\\ 12)\left\{\begin{array}{cc} \dfrac{1-\sqrt{x}}{\sqrt{x}+1}, & 0 \le x <1\\ \dfrac{\sqrt{x}-1}{\sqrt{x}+1}, & x \ge 1 \end{array} \right.$
Giải thích các bước giải:
$7)\\ \dfrac{\sqrt{45mn^2}}{\sqrt{20m}} \ \ \ \ m,n>0\\ =\dfrac{\sqrt{9.5mn^2}}{\sqrt{4.5m}}\\ =\dfrac{3\sqrt{5m}|n|}{2\sqrt{5m}} \\ =\dfrac{3n}{2} \\ 8)\\ \dfrac{\sqrt{16a^4b^6}}{\sqrt{128a^6b^6}}\ \ \ \ a<0; b \ne 0\\ =\dfrac{\sqrt{4^2a^4b^6}}{\sqrt{64.2a^6b^6}}\\ =\dfrac{4|a^2b^3|}{8\sqrt{2}|a^3b^3|}\\ =\dfrac{4a^2|b^3|}{8\sqrt{2}.(-a^3)|b^3|}\\ =-\dfrac{1}{2\sqrt{2}a}\\ =-\dfrac{\sqrt{2}}{4a}\\ 9)\\ ab^2\sqrt{\dfrac{3}{a^2b^4}}\ \ \ \ a<0; b \ne 0\\ =ab^2\dfrac{\sqrt{3}}{\sqrt{a^2b^4}}\\ =ab^2\dfrac{\sqrt{3}}{|ab^2|}\\ =ab^2\dfrac{\sqrt{3}}{(-a)b^2}\\ =-\sqrt{3}\\ 10)\\ \sqrt{\dfrac{27(a-3)^2}{48}} \ \ \ \ a>3\\ =\sqrt{\dfrac{9(a-3)^2}{16}} \\ =\sqrt{\dfrac{(3(a-3))^2}{4^2}} \\ =\dfrac{|3(a-3)|}{4} \\ =\dfrac{3(a-3)}{4} \\ 11)\\ \sqrt{\dfrac{9+12a+4a^2}{b^2}} \ \ \ \ a \ge -1,5; b<0\\ =\sqrt{\dfrac{4a^2+12a+9}{b^2}}\\ =\sqrt{\dfrac{(2a)^2+2.2a.3+3^2}{b^2}}\\ =\sqrt{\dfrac{(2a+3)^2}{b^2}}\\ =\dfrac{|2a+3|}{|b|}\\ =-\dfrac{2a+3}{b}\\ 12)\\ \sqrt{\dfrac{x-2\sqrt{x}+1}{x+2\sqrt{x}+1}} \ \ \ \ x \ge 0\\ =\sqrt{\dfrac{(\sqrt{x}-1)^2}{(\sqrt{x}+1)^2}}\\ =\dfrac{|\sqrt{x}-1|}{|\sqrt{x}+1|}\\ =\left\{\begin{array}{cc} \dfrac{1-\sqrt{x}}{\sqrt{x}+1}, & 0 \le x <1\\ \dfrac{\sqrt{x}-1}{\sqrt{x}+1}, & x \ge 1 \end{array} \right.$
