Question

Answer 1

1.

\(\sqrt{10+2\sqrt{21}}-\sqrt{10-2\sqrt{21}}=\sqrt{7+3+2\sqrt{7.3}}-\sqrt{7+3-2\sqrt{7.3}}\\ =\sqrt{(\sqrt{7}+\sqrt{3})^2}-\sqrt{(\sqrt{7}-\sqrt{3})^2}\\ =|\sqrt{7}+\sqrt{3}|-|\sqrt{7}-\sqrt{3}|\\ =\sqrt{7}+\sqrt{3}-(\sqrt{7}-\sqrt{3})\\ =2\sqrt{3}\)

2.

\(=\sqrt{16+2\sqrt{48}}-|\sqrt{12}-1|\\ =\sqrt{12+4+2\sqrt{12.4}}-(\sqrt{12}-1)\\ =\sqrt{(\sqrt{12}+\sqrt{4})^2}-(\sqrt{12}-1)\\ =|\sqrt{12}+\sqrt{4}|-(\sqrt{12}-1)\\ =\sqrt{12}+\sqrt{4}-\sqrt{12}+1=\sqrt{4}+1=2+1=3\)

Answer 2

3.

\(=|3-\sqrt{5}|-\sqrt{5+4+2\sqrt{4.5}}\\ =(3-\sqrt{5})-\sqrt{(\sqrt{5}+\sqrt{4})^2}\\ =(3-\sqrt{5})-|\sqrt{5}+\sqrt{4}|\\ =(3-\sqrt{5})-(\sqrt{5}+\sqrt{4})\\ =3-\sqrt{4}=3-2=1\)

4.

\(=\sqrt{12+9-2\sqrt{12.9}}-|1-2\sqrt{3}|\\ =\sqrt{(\sqrt{12}-\sqrt{9})^2}-(2\sqrt{3}-1)\\ =|\sqrt{12}-\sqrt{9}|-(2\sqrt{3}-1)\\ =\sqrt{12}-\sqrt{9}-2\sqrt{3}+1\\ =2\sqrt{3}-3-2\sqrt{3}+1=-3+1=-2\)

Answer 3

5.

\(=|3\sqrt{3}-2|+\sqrt{4+3-2\sqrt{4.3}}\\ =(3\sqrt{3}-2)+\sqrt{(\sqrt{4}-\sqrt{3})^2}\\ =(3\sqrt{3}-2)+|\sqrt{4}-\sqrt{3}|\\ =3\sqrt{3}-2+\sqrt{4}-\sqrt{3}=2\sqrt{3}-2+2=2\sqrt{3}\)

6.

\(=|\sqrt{3}-\sqrt{5}|-\sqrt{20+3-2\sqrt{20.3}}\\ =\sqrt{5}-\sqrt{3}-\sqrt{(\sqrt{20}-\sqrt{3})^2}\\ =\sqrt{5}-\sqrt{3}-(\sqrt{20}-\sqrt{3})\\ =\sqrt{5}-\sqrt{3}-\sqrt{20}+\sqrt{3}=\sqrt{5}-\sqrt{20}\\ =\sqrt{5}-2\sqrt{5}=-\sqrt{5}\)

Answer 4

7.

\(=\sqrt{16+2-2\sqrt{16.2}}+\sqrt{25+2-2\sqrt{25.2}}\\ =\sqrt{(\sqrt{16}-\sqrt{2})^2}+\sqrt{(\sqrt{25}-\sqrt{2})^2}\\ =|\sqrt{16}-\sqrt{2}|+|\sqrt{25}-\sqrt{2}|\\ =\sqrt{16}-\sqrt{2}+\sqrt{25}-\sqrt{2}=4-2\sqrt{2}+5\\ =9-2\sqrt{2}\)

Answer 5

8.

\(=2.3\sqrt{5}-7\sqrt{5}+\sqrt{5+4+2\sqrt{5.4}}\\ =6\sqrt{5}-7\sqrt{5}+\sqrt{(\sqrt{5}+\sqrt{4})^2}\\ =-\sqrt{5}+|\sqrt{5}+\sqrt{4}|\\ =-\sqrt{5}+\sqrt{5}+\sqrt{4}=\sqrt{4}=2\)

Answer 6

 

10.

\(=\sqrt{7+4+2\sqrt{7.4}}-\sqrt{25+7-2\sqrt{25.7}}\\ =\sqrt{(\sqrt{7}+\sqrt{4})^2}-\sqrt{(\sqrt{25}-\sqrt{7})^2}\\ =\sqrt{7}+\sqrt{4}-(\sqrt{25}-\sqrt{7})\\ =\sqrt{7}+\sqrt{4}-\sqrt{25}+\sqrt{7}=2\sqrt{7}+2-5=2\sqrt{7}-3\)

Answer 7

9.

\(=\sqrt{9+5-2\sqrt{9.5}}+|3\sqrt{5}-7|\\ =\sqrt{(\sqrt{9}-\sqrt{5})^2}+(7-3\sqrt{5})\\ =\sqrt{9}-\sqrt{5}+7-3\sqrt{5}\\ =3+7-4\sqrt{5}=10-4\sqrt{5}\)

Answer 8

Cách để khai căn biểu thức $\sqrt{(a+2b\sqrt{c}}$
Để khai căn, bạn dự định sẽ biến đổi $a+2b\sqrt{c}=a+2\sqrt{b^2c}=(\sqrt{m}+\sqrt{n})^2$

$\Leftrightarrow a+2\sqrt{b^2c}=m+n+2\sqrt{mn}$
Cho $m+n=a; mn=b^2c$ trong đó $a,b,c$ là những số đã biết rồi.

Theo định lý Viet đảo, để tìm $m,n$ bạn chỉ cần giải PT $X^2-aX+b^2c=0$ là được.

Khi đó bạn sẽ tìm ra $m,n$ phù hợp để phá căn.