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

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

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

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

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

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

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

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

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

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

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 !!!

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

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

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

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

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

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

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

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

\(A=-3\)

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

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

\(B=1\)

1) \(\sqrt{9+4\sqrt{5}}-\sqrt{9-4\sqrt{5}}\)

\(=\sqrt{2^2+2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}-\sqrt{2^2-2\cdot2\cdot\sqrt{5}+\left(\sqrt{5}\right)^2}\)

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

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

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

\(=4\)

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

\(=\sqrt{3^2-2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}+\sqrt{3^2+2\cdot3\cdot\sqrt{3}+\left(\sqrt{3}\right)^2}\)

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

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

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

\(=6\)

Ta có : 

\(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)

\(=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-\sqrt{128}}}}\)

Ta có : 

\(18-\sqrt{128}=18-8\sqrt{2}=16-2.4.\sqrt{2}+2=\left(4-\sqrt{2}\right)^2\)

Vậy 

\(\sqrt{18-\sqrt{128}}=4-\sqrt{2}\)

Thay vào ta có

\(\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{18-\sqrt{128}}}}\)

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

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

Lại có : 

\(4+2\sqrt{3}=3+2.1.\sqrt{3}+1=\left(\sqrt{3}+1\right)^2\)

Do đó : 

\(\sqrt{4+2\sqrt{3}}=\sqrt{\left(\sqrt{3}+1\right)^2}=\sqrt{3}+1\)

Vậy : 

\(\sqrt{6-2\sqrt{4+2\sqrt{3}}}=\sqrt{6-2\left(\sqrt{3}+1\right)}\)

\(=\sqrt{4-2\sqrt{3}}\)

\(=\sqrt{3-2.1.\sqrt{3}+1}\)

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

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

Vậy : \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}=\sqrt{3}-1\)

\(\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)

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

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

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

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

\(=5\)

9: \(A=\dfrac{\sqrt{8+2\sqrt{15}}-\sqrt{14-6\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}+\sqrt{3}-3+\sqrt{5}}{\sqrt{2}}=\dfrac{2\sqrt{10}+\sqrt{6}-3\sqrt{2}}{2}\)

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

\(=\dfrac{\sqrt{3}+1+\sqrt{3}-1}{\sqrt{2}}=\dfrac{2\sqrt{3}}{\sqrt{2}}=\sqrt{6}\)

11: \(A=\dfrac{\sqrt{24-6\sqrt{7}}-\sqrt{24+6\sqrt{7}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{21}-\sqrt{3}-\sqrt{21}-\sqrt{3}}{\sqrt{2}}=-\dfrac{2\sqrt{3}}{\sqrt{2}}=-\sqrt{6}\)

12: \(B=\left(3+\sqrt{3}\right)\sqrt{12-6\sqrt{3}}\)

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

=9-3=6

13: \(A=\sqrt{5}-2-\left(3-\sqrt{5}\right)\)

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

\(A=\sqrt{\left(9\sqrt{2}+2\sqrt{3}\right)^2}-\sqrt{\left(9\sqrt{2}-\sqrt{3}\right)^2}\)

\(=\left|9\sqrt{2}+2\sqrt{3}\right|-\left|9\sqrt{2}-\sqrt{3}\right|\)

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

Kiểm tra lại đề bài câu B, chỗ \(\sqrt{2+\sqrt{2+2}}\)

Câu B đúng đề bài ạ ! 

30,001x3​=3(0,1x)3​=0,1x;

\sqrt[3]{-125 a^{12}}=\sqrt[3]{\left(-5 a^{4}\right)^{3}}=-5 a^{4};3−125a12​=3(−5a4)3​=−5a4;

\sqrt[3]{27 x^{6}}=\sqrt[3]{\left(3 x^{2}\right)^{3}}=3 x^{2};327x6​=3(3x2)3​=3x2;

\sqrt[3]{-0,343 a^{3}}=\sqrt[3]{(-0,7 a)^{3}}=-0,7 a;3−0,343a3​=3(−0,7a)3​=−0,7a;

Ta rút gọn các biểu thức như sau:

\(\sqrt[3]{0,001x^3}=\sqrt[3]{\left(0,1x\right)^3}=0,1x.\)

\(\sqrt[3]{-125a^{12}}=\sqrt[3]{\left(-5a^4\right)^3}=-5a^4\)

\(\sqrt[3]{27x^6}=\sqrt[3]{\left(3x^2\right)^3}=3x^2\)

\(\sqrt[3]{-0,343a^3}=\sqrt[3]{\left(-0,7a\right)^3}=-0,7a\)

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

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

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

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