\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

\(\sqrt{3^2-2.3.\sqrt{6}+6}+\sqrt{3^2-2.3.2\sqrt{6}+\left(2.\sqrt{6}\right)^2}\)

= \(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(3-2.\sqrt{6}\right)^2}=\left|3-\sqrt{6}\right|+\left|3-2\sqrt{6}\right|=3-\sqrt{6}-3+2\sqrt{6}=\sqrt{6}\)

1. \(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{5}+\sqrt{2}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\\ =\sqrt{5}+\sqrt{2}-\sqrt{5}+\sqrt{2}=2\sqrt{2}\)

2. \(\sqrt{12-6\sqrt{3}}+\sqrt{21-12\sqrt{3}}=\sqrt{\left(3-\sqrt{3}\right)^2}+\sqrt{\left(2\sqrt{3}-3\right)^2}\\ =3-\sqrt{3}+2\sqrt{3}-3=\sqrt{3}\)

3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(3+\sqrt{6}\right)^2}\\ =2\sqrt{6}-3+3+\sqrt{6}=3\sqrt{6}\)

1.\(\sqrt{7+2\sqrt{10}}-\sqrt{7-2\sqrt{10}}=\sqrt{\left(\sqrt{2}+\sqrt{5}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{2}\right)^2}\)

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

2. \(\sqrt{12-6\sqrt{3}+\sqrt{21-12\sqrt{3}}}=\sqrt{12-6\sqrt{3}+\sqrt{\left(3-2\sqrt{3}\right)^2}}\)

\(=\sqrt{12-6\sqrt{3}+2\sqrt{3}-3}=\sqrt{9-4\sqrt{3}}\)

3. \(\sqrt{33-12\sqrt{6}}+\sqrt{15-6\sqrt{6}}=\sqrt{\left(2\sqrt{6}-3\right)^2}+\sqrt{\left(\sqrt{6}-3\right)^2}\)

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

\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-2.3.\sqrt{6}}+\sqrt{33-2.6\sqrt{6}}\)

\(=\sqrt{3^2-2.3.\sqrt{6}+\sqrt{6^2}}+\sqrt{24-2.2\sqrt{6}.3+9}\)

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

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

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

\(=6-3\sqrt{6}\)

Ko vt lại đề nha bn:

\(=\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

\(=\sqrt{15-2.3.\sqrt{6}}+\sqrt{33-2.6\sqrt{6}}\)

\(=\sqrt{3^2-2.3.\sqrt{6}+\sqrt{6^2}}+\sqrt{24-2.2\sqrt{6}.3+9}\)

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

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

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

\(=6-3\sqrt{6}\)

Rất vui vì giúp đc bn !!!

\(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}=2-\sqrt{3}+\sqrt{3-2\sqrt{3}+1}\left(\text{vì }2>\sqrt{3}\right)\)

\(=2-\sqrt{3}+\sqrt{\left(\sqrt{3}-1\right)^2}=2-\sqrt{3}+\sqrt{3}-1\left(\text{vì }\sqrt{3}>1\right)\)

\(=1\)

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{\left(2\sqrt{6}\right)^2-2.2\sqrt{6}.3+9}\)

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3\left(\text{vì }3>\sqrt{6};2\sqrt{6}>3\right)\)

\(=\sqrt{6}\)

a/

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

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

\(=1\)

b/

\(=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(\sqrt{24}-3\right)^2}\)

\(=3-\sqrt{6}+\sqrt{24}-3\)

\(=-\sqrt{6}+2\sqrt{6}=\sqrt{6}\)

tick cho mình nha

a) \(\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{4-2\sqrt{3}}\)

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

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

b) \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

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

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

c) \(\left(15\sqrt{200}-3\sqrt{450}+2\sqrt{50}\right):\sqrt{10}\)

= \(\dfrac{15\sqrt{200}}{\sqrt{10}}-\dfrac{3\sqrt{450}}{\sqrt{10}}+\dfrac{2\sqrt{50}}{\sqrt{10}}\)

= \(15\sqrt{20}-3\sqrt{45}+2\sqrt{5}\)

= \(30\sqrt{5}-9\sqrt{5}+2\sqrt{5}\) = \(23\sqrt{5}\)

Câu 1:

a: \(\dfrac{2}{5}\sqrt{75}-0,5\cdot\sqrt{48}+\sqrt{300}-\dfrac{2}{3}\cdot\sqrt{12}\)

\(=\dfrac{2}{5}\cdot5\sqrt{3}-0,5\cdot4\sqrt{3}+10\sqrt{3}-\dfrac{2}{3}\cdot2\sqrt{3}\)

\(=2\sqrt{3}-2\sqrt{3}+10\sqrt{3}-\dfrac{4}{3}\sqrt{3}\)

\(=10\sqrt{3}-\dfrac{4}{3}\sqrt{3}=\dfrac{26}{3}\sqrt{3}\)

b: \(\dfrac{9-2\sqrt{3}}{3\sqrt{6}-2\sqrt{2}}+\dfrac{3}{3+\sqrt{6}}\)

\(=\dfrac{\sqrt{3}\cdot3\sqrt{3}-2\sqrt{3}}{\sqrt{2}\left(3\sqrt{3}-2\right)}+\dfrac{3\left(3-\sqrt{6}\right)}{9-6}\)

\(=\dfrac{\sqrt{3}\left(3\sqrt{3}-2\right)}{\sqrt{2}\left(3\sqrt{3}-2\right)}+3-\sqrt{6}\)

\(=\dfrac{\sqrt{3}}{\sqrt{2}}+3-\sqrt{6}=3-\dfrac{\sqrt{6}}{2}\)

c: \(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}\)

=\(\sqrt{9-2\cdot3\cdot\sqrt{6}+6}+\sqrt{24-2\cdot2\sqrt{6}\cdot3+9}\)

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

\(=\left|3-\sqrt{6}\right|+\left|2\sqrt{6}-3\right|\)

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

Bài 2:

a: 

b: Phương trình hoành độ giao điểm là:

\(3x+2=-x-4\)

=>4x=-6

=>x=-3/2

Thay x=-3/2 vào y=-x-4, ta được:

\(y=-\left(-\dfrac{3}{2}\right)-4=\dfrac{3}{2}-4=-\dfrac{5}{2}\)

Vậy: \(A\left(-\dfrac{3}{2};-\dfrac{5}{2}\right)\)

c: Vì (d2)//(d) nên \(\left\{{}\begin{matrix}a=-1\\b\ne-4\end{matrix}\right.\)

Vậy: (d2): y=-x+b

Thay x=-2 và y=5 vào (d2), ta được:

\(b-\left(-2\right)=5\)

=>b+2=5

=>b=5-2=3

Vậy: (d2): y=-x+3

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{9-2.3\sqrt{6}+6}+\sqrt{24-2.2\sqrt{6}.3+9}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

\(\sqrt{15-6\sqrt{6}}+\sqrt{33-12\sqrt{6}}=\sqrt{\left(3-\sqrt{6}\right)^2}+\sqrt{\left(2\sqrt{6}-3\right)^2}=3-\sqrt{6}+2\sqrt{6}-3=\sqrt{6}\)

\(\dfrac{\sqrt{15}-\sqrt{6}}{\sqrt{35}-\sqrt{14}}=\dfrac{\sqrt{3}\left(\sqrt{5}-\sqrt{2}\right)}{\sqrt{7}\left(\sqrt{5}-\sqrt{2}\right)}=\sqrt{\dfrac{3}{7}}\)

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

\(\dfrac{2\sqrt{8}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}=\dfrac{2\left(\sqrt{8}-\sqrt{3}\right)}{\sqrt{6}\left(\sqrt{3}-\sqrt{8}\right)}=-\dfrac{2\sqrt{6}}{6}\)

`(sqrt 15 - sqrt 6)/(sqrt 35 - sqrt 14)`

`= (sqrt 3 . (sqrt 5 - sqrt 2))/(sqrt 7. (sqrt 5 - sqrt 2))`

`= sqrt3/sqrt 7`

`@ (sqrt 15 - sqrt 5)/(sqrt 3 - 1)`

`= (sqrt 5(sqrt 3 - 1))/(sqrt 3 - 1)`

`= sqrt5`

`@ (2 sqrt 8 - sqrt 12)/(sqrt18 - sqrt 48)`

`= (2(sqrt 8 - sqrt 3)/(sqrt 6(sqrt 3 - sqrt 8))`

`= (-2)/(sqrt 6) = (-2 sqrt 6)/6`