1, \(\sqrt{8+2\sqrt{15}}=\sqrt{8+2\sqrt{5.3}}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)

2, \(\sqrt{15-2\sqrt{14}}=\sqrt{14-2\sqrt{14}+1}=\sqrt{\left(\sqrt{14}-1\right)^2}=\sqrt{14}-1\)

3, \(\sqrt{21+8\sqrt{5}}=\sqrt{21+2.4\sqrt{5}}=\sqrt{16+2.4\sqrt{5}+5}\)

\(=\sqrt{\left(4+\sqrt{5}\right)^2}=4+\sqrt{5}\)

1.\(=5\sqrt{2}-3\sqrt{2}+10\sqrt{2}-9\sqrt{2}=3\sqrt{2}\)

2.\(=5\sqrt{5}+4\sqrt{5}-9\sqrt{5}=0\)

a ) \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)

b ) \(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)

c ) \(\sqrt{21+4\sqrt{5}}=\sqrt{\left(2\sqrt{5}+1\right)^2}=2\sqrt{5}+1\)

d ) \(\sqrt{11+4\sqrt{7}}=\sqrt{\left(\sqrt{7}+2\right)^2}=\sqrt{7}+2\)

a) \(\sqrt{5-2\sqrt{6}}=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)

b) \(\sqrt{3-2\sqrt{2}}=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)

c) \(\sqrt{21+4\sqrt{5}}=\sqrt{\left(2\sqrt{5}+1\right)^2}=2\sqrt{5}+1\)

d) \(\sqrt{11+4\sqrt{7}}=\sqrt{\left(\sqrt{7}+2\right)^2}=\sqrt{7}+2\)

a

\(\sqrt{5-2\sqrt{6}}=\sqrt{\sqrt{2}^2-2.\sqrt{2}.\sqrt{3}+\sqrt{3}^2}\)

\(=\sqrt{\left(\sqrt{3}-\sqrt{2}\right)^2}=\sqrt{3}-\sqrt{2}\)

b,

\(\sqrt{3-2\sqrt{2}}=\sqrt{\sqrt{2}^2-2.\sqrt{2}.1+1^2}\)

\(=\sqrt{\left(\sqrt{2}-1\right)^2}=\sqrt{2}-1\)

c,\(\sqrt{11+4\sqrt{7}}=\sqrt{11+2\sqrt{28}}=\sqrt{\sqrt{7}^2+2.\sqrt{7}.\sqrt{4}+\sqrt{4}^2}\)

\(=\sqrt{\left(\sqrt{7}+\sqrt{4}\right)^2}=\sqrt{7}+\sqrt{4}\)

giúp mình với ạ !

Mình làm 5 bài trắc nha

Bài 1: Đưa thừa số ra ngoài dấu căn:

\(2\sqrt{225a^2}=2.15a=30a\)

Bài 2: Đưa thừa số vào trong dấu căn :

\(x\sqrt{\dfrac{-39}{x}}=\sqrt{x^2.\dfrac{-39}{x}}=\sqrt{-39x}\)

Bài 3: Sắp xếp theo thứ tự tăng dần :

a) \(2\sqrt{3}< 3\sqrt{2}< 2\sqrt{5}< 5\sqrt{2}\)

b) \(4\sqrt{2}< \sqrt{37}< 2\sqrt{15}< 3\sqrt{7}\)

c) \(6\sqrt{\dfrac{1}{3}}< \sqrt{27}< 2\sqrt{28}< 5\sqrt{7}\)

a, \(\sqrt{5\left(1-\sqrt{2}\right)^2}=\sqrt{5}.\sqrt{\left(1-\sqrt{2}\right)^2}\)

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

b, \(\sqrt{27\left(2-\sqrt{5}\right)^2}=\sqrt{27}.\sqrt{\left(2-\sqrt{5}\right)^2}\)

\(=\sqrt{27}.\left(2-\sqrt{5}\right)=2\sqrt{27}-\sqrt{135}\)

c, \(\sqrt{\dfrac{2}{\left(3-\sqrt{10}\right)^2}}=\dfrac{\sqrt{2}}{\sqrt{\left(3-\sqrt{10}\right)^2}}\)

\(=\dfrac{\sqrt{2}}{3-\sqrt{10}}\)

d, \(\sqrt{\dfrac{5\left(1-\sqrt{3}\right)^2}{4}}=\dfrac{\sqrt{5\left(1-\sqrt{3}\right)^2}}{\sqrt{4}}\)

\(=\dfrac{\sqrt{5}.\left(1-\sqrt{3}\right)}{2}=\dfrac{\sqrt{5}-\sqrt{15}}{2}\)

Chúc bạn học tốt!!!

a) \(\sqrt{5\left(1-\sqrt{2}\right)^2}\)

= \(\sqrt{5}.\sqrt{\left(1-\sqrt{2}\right)^2}\)

= \(\sqrt{5}.\left(\sqrt{2}-1\right)\)

= \(\sqrt{10}-\sqrt{5}\)

b) \(\sqrt{27\left(2-\sqrt{5}\right)^2}\)

= \(\sqrt{27}.\sqrt{\left(2-\sqrt{5}\right)^2}\)

= \(\sqrt{27}.\left(\sqrt{5}-2\right)\)

= \(\sqrt{135}-2\sqrt{27}\)

c) \(\sqrt{\dfrac{2}{\left(3-\sqrt{10}\right)^2}}\)

= \(\dfrac{\sqrt{2}}{\sqrt{\left(3-\sqrt{10}\right)^2}}\)

= \(\dfrac{\sqrt{2}}{\sqrt{10}-3}\)

d) \(\sqrt{\dfrac{5\left(1-\sqrt{3}\right)^2}{4}}\)

= \(\dfrac{\sqrt{5}.\sqrt{\left(1-\sqrt{3}\right)^2}}{\sqrt{4}}\)

= \(\dfrac{\sqrt{5}.\left(\sqrt{3}-1\right)}{2}\)

= \(\dfrac{\sqrt{15}-\sqrt{5}}{2}\)

Câu 3:

\(=3\sqrt{3}+8\sqrt{3}-15\sqrt{3}=-4\sqrt{3}\)

Câu 6:

=>3x=9 và x+y=4

=>x=3 và y=1