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

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

\(\sqrt{19+8\sqrt{3}}=\sqrt{16+8\sqrt{3}+3}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}\)

\(=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=\sqrt{9-6\sqrt{5}+5}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}\)

\(=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

\(\sqrt{12-6\sqrt{3}}=\sqrt{3^2-2.3.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(3-\sqrt{3}\right)^2}=\left|3-\sqrt{3}\right|=3-\sqrt{3}\)

\(\sqrt{19+8\sqrt{3}}=\sqrt{4^2+2.4.\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(4+\sqrt{3}\right)^2}=\left|4+\sqrt{3}\right|=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=\sqrt{3^2-2.3.\sqrt{5}+\left(\sqrt{5}\right)^2}=\sqrt{\left(3-\sqrt{5}\right)^2}=\left|3-\sqrt{5}\right|=3-\sqrt{5}\)

\(\sqrt{12-6\sqrt{3}}=3-\sqrt{3}\)

\(\sqrt{19+8\sqrt{3}}=4+\sqrt{3}\)

\(\sqrt{14-6\sqrt{5}}=3-\sqrt{5}\)

Đặt y = \(x+1=\sqrt[3]{8+2\sqrt{14}}+\sqrt[3]{8-2\sqrt{14}}\)

=> \(y^3=8+2\sqrt{14}+8-2\sqrt{14}+3\sqrt[3]{\left(8+2\sqrt{14}\right)\left(8-2\sqrt{14}\right)}.y\)

<=> \(y^3=16+6y\)

=> \(\left(x+1\right)^3=16+6\left(x+1\right)\)

=> \(x^3+3x^2+3x+1=6x+32\)

<=> \(x^3+3x^2-3x-5=26\)

Ta có: 

\(x^6+3x^5-3x^4-2x^3+9x^2-9x+2018\)

= \(x^6+3x^5-3x^4-5x^3+3x^3+9x^2-9x-15+2033\)

= \(\left(x^3+3x^2-3x-5\right)\left(x^3+3\right)+2033\)

= \(26x^3+2111\)

\(=26\left(\sqrt[8]{8+2\sqrt{14}}+\sqrt[8]{8-2\sqrt{14}}-1\right)^3+2033\)

Bài làm:

\(\sqrt{14-8\sqrt{3}}-\sqrt{24-12\sqrt{3}}\)

\(=\sqrt{8-8\sqrt{3}+6}-\sqrt{18-12\sqrt{3}+6}\)

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

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

\(=-\sqrt{2}\)

Học tốt 

\(\sqrt{3.2-8\sqrt{3}+4.2}=\sqrt{\left(\sqrt{3}.\sqrt{2}\right)^2-2.2.\sqrt{2}.\sqrt{2}.\sqrt{3}+\left(2\sqrt{2}\right)^2}\)

\(=\sqrt{\left(2\sqrt{2}-\sqrt{6}\right)^2}-\sqrt{3^2.2-12\sqrt{3}+3.2}=2\sqrt{2}-\sqrt{6}-3\sqrt{2}+\sqrt{6}=-\sqrt{2}\)

\(A=\sqrt{\frac{1}{8+3\sqrt{7}}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(A=\sqrt{\frac{2}{16+6\sqrt{7}}}\left(3\sqrt{2}+\sqrt{14}\right)\)
\(A=\frac{\sqrt{2}}{3+\sqrt{7}}\left(3+\sqrt{7}\right)\sqrt{2}\)
\(A=2\)

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

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

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

5) \(=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}\)

6) \(=\sqrt{\left(\sqrt{7}-\sqrt{3}\right)^2}=\sqrt{7}-\sqrt{3}\)

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

Thêm câu này hộ tớ nx nhé !
e) \(\left(\sqrt{8}-3\sqrt{2}+\sqrt{10}\right).\left(\sqrt{2}-3\sqrt{0.4}\right)\)

\(a,\left(\frac{2\sqrt{3}-\sqrt{6}}{\sqrt{8}-2}-\frac{\sqrt{216}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{12}-\sqrt{6}}{2\left(\sqrt{2}-1\right)}-\frac{6\sqrt{6}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}\left(\sqrt{2}-1\right)}{2\left(\sqrt{2}-1\right)}-2\sqrt{6}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{\sqrt{6}}{2}-\frac{4\sqrt{6}}{2}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{\sqrt{6}-4\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{-3\sqrt{6}}{2}\cdot\frac{1}{\sqrt{6}}\)

\(=-\frac{3}{2}\)

\(b,\left(\frac{\sqrt{14}-\sqrt{7}}{1-\sqrt{2}}+\frac{\sqrt{15}-\sqrt{5}}{1-\sqrt{3}}\right):\frac{1}{\sqrt{7}-\sqrt{5}}\)

\(=\left(\frac{\sqrt{7}\left(\sqrt{2}-1\right)}{1-\sqrt{2}}+\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{1-\sqrt{3}}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(\frac{-\sqrt{7}\left(1-\sqrt{2}\right)}{1-\sqrt{2}}-\frac{\sqrt{5}\left(\sqrt{3}-1\right)}{\sqrt{3}-1}\right).\left(\sqrt{7}-\sqrt{5}\right)\)

\(=\left(-\sqrt{7}-\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(\sqrt{7}+\sqrt{5}\right)\left(\sqrt{7}-\sqrt{5}\right)\)

\(=-\left(7-5\right)\)

\(=-2\)