\(A=\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}=\sqrt{\left(4+\sqrt{8}\right)^2}-\sqrt{\left(4-\sqrt{8}\right)^2}=\left|4+\sqrt{8}\right|-\left|4-\sqrt{8}\right|=4+\sqrt{8}-4+\sqrt{8}=4\sqrt{2}\)

\(A=\sqrt{24+16\sqrt{2}}-\sqrt{24-16\sqrt{2}}\)

\(=\sqrt{8+2.4.2\sqrt{2}+16}-\sqrt{16-2.4.2\sqrt{2}+8}\)

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

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

\(=4\sqrt{2}\)

Ta có: \(x=\sqrt{97-56\sqrt{3}}+\sqrt{52+16\sqrt{3}}\)

\(=\sqrt{49-2\cdot7\cdot4\sqrt{3}+48}+\sqrt{48+2\cdot4\sqrt{3}\cdot2+4}\)

\(=\sqrt{\left(7-4\sqrt{3}\right)^2}+\sqrt{\left(4\sqrt{3}+2\right)^2}\)

\(=\left|7-4\sqrt{3}\right|+\left|4\sqrt{3}+2\right|\)

\(=7-4\sqrt{3}+4\sqrt{3}+2\)

\(=9\)

Làm luôn phần y :D

y = \(\sqrt{33+20\sqrt{2}}+\sqrt{24-16\sqrt{2}}\)

y = \(\sqrt{33+2.10\sqrt{2}}+\sqrt{24-2.8\sqrt{2}}\)

y = \(\sqrt{33+2.5.2\sqrt{2}}+\sqrt{24-2.4.2\sqrt{2}}\)

y = \(\sqrt{25+2.5.\sqrt{8}+8}+\sqrt{16-2.4.\sqrt{8}+8}\)

y = \(\sqrt{\left(5+\sqrt{8}\right)^2}+\sqrt{\left(4-\sqrt{8}\right)^2}\)

y = |5 + \(\sqrt{8}\)| + |4 - \(\sqrt{8}\)| 

y = 5 + \(\sqrt{8}\) + 4 - \(\sqrt{8}\)   (Vì 4 > \(\sqrt{8}\) nên 4 - \(\sqrt{8}\) > 0)

y = 9

Vậy y = 9

Chúc bn học tốt!

\(x^3=\left(\sqrt[3]{16-\sqrt{255}}+\dfrac{1}{\sqrt[3]{16-\sqrt{255}}}\right)^3\)

=>\(x^3=16-\sqrt{255}+\dfrac{1}{16-\sqrt{255}}+3\cdot x\cdot\sqrt[3]{\left(16-\sqrt{255}\right)\cdot\dfrac{1}{16-\sqrt{255}}}\)

=>\(x^3=16-\sqrt{255}+\dfrac{1}{16-\sqrt{255}}+3x\)

=>\(x^3-3x=16-\sqrt{255}+\dfrac{1}{16-\sqrt{255}}\)

=>\(B=16-\sqrt{255}+\dfrac{1}{16-\sqrt{255}}+1077\)

\(=1093-\sqrt{255}+\dfrac{16+\sqrt{255}}{1}\)

\(=1093-\sqrt{255}+16+\sqrt{255}\)

=1093+16

=1109

\(\left(\frac{2}{5}\sqrt{16}+2\sqrt{\frac{16}{25}}\right):2\sqrt{\frac{1}{16}}=\left(\frac{2}{5}.\sqrt{4^2}+2\sqrt{\frac{4^2}{5^2}}\right):\frac{2}{\sqrt{4^2}}\)

\(=\left(\frac{2}{5}.4+2.\frac{4}{5}\right).2=\left(\frac{8}{5}+\frac{8}{5}\right).2=\frac{32}{5}\)

\(\left(\frac{2}{5}.\sqrt{16}+2\sqrt{\frac{16}{25}}\right):2\sqrt{\frac{1}{16}}\)

\(=\left(\frac{2}{5}.4+2.\frac{4}{5}\right):2.\frac{1}{4}\)

\(=\left(\frac{8}{5}+\frac{8}{5}\right):\frac{1}{2}\)

\(=\frac{16}{5}:\frac{1}{2}\)

\(=\frac{32}{5}\)

 ^...^ ^_^

\(\left(\frac{2}{5}\sqrt{16}+2\sqrt{\frac{16}{25}}\right):2\sqrt{\frac{1}{16}}\)

\(=\left(\frac{2}{5}.4+2.\frac{4}{5}\right):2.\frac{1}{4}\)

\(=\left(\frac{8}{5}+\frac{8}{5}\right):\frac{1}{2}\)

\(=\frac{16}{5}:\frac{1}{2}\)

\(=\frac{16}{5}.2\)

\(=\frac{32}{5}\)

giải chi tiết hộ mk nhá

\(\sqrt[3]{16-8\sqrt{5}}\)=\(\sqrt[3]{1-3\sqrt{5}+15-5\sqrt{5}}\)=\(\sqrt[3]{1-3\sqrt{5}+3\left(\sqrt{5}\right)^2-\left(\sqrt{5}\right)^3}\)=\(\sqrt[3]{\left(1-\sqrt{5}\right)^3}\)=\(1-\sqrt{5}\)

làm tương tự: \(\sqrt[3]{16+8\sqrt{5}}\)=\(1+\sqrt{5}\)

suy ra: a = 2

Lời giải:

\(\left(\frac{15}{\sqrt{6}+1}+\frac{4}{\sqrt{16}-2}-\frac{12}{3-\sqrt{16}}\right).(\sqrt{6}+11)=\left(\frac{15(\sqrt{6}-1)}{(\sqrt{6}+1)(\sqrt{6}-1)}+\frac{4}{4-2}-\frac{12}{3-4}\right)(\sqrt{6}+11)\)

\(=\left(\frac{15(\sqrt{6}-1)}{6-1}+2+12\right)(\sqrt{6}+11)=(3\sqrt{6}-3+14)(\sqrt{6}+11)\)

\(=(3\sqrt{6}+11)(\sqrt{6}+11)\)

\(=\dfrac{\sqrt{26}}{\sqrt{676.26}}.\dfrac{\sqrt{6}.4}{\sqrt{2}}=\dfrac{\sqrt{26}}{\sqrt{676}.\sqrt{26}}.\dfrac{\sqrt{12}.4}{2}=\dfrac{4\sqrt{3}}{26}=\dfrac{2\sqrt{3}}{13}\)

169*2097152^(1/4)*căn bậc hai(6)

a) Vì \({4^2} = 16\) nên \(\sqrt {16}  = 4\)

b) Vì \({9^2} = 81\) nên \(\sqrt {81}  = 9\)

c) Vì 2021 > 0 nên \(\sqrt {{{2021}^2}}  = 2021\)

\(a,\sqrt{42}=\sqrt{3\cdot14}>\sqrt{3\cdot12}=6\\ \sqrt[3]{51}=\sqrt[3]{17}< \sqrt[3]{3\cdot72}=6\\ \Rightarrow\sqrt{42}>\sqrt[3]{51}\\ b,16^{\sqrt{3}}=4^{2\sqrt{3}}\\ 18>12\Rightarrow3\sqrt{2}>2\sqrt{3}\Rightarrow4^{3\sqrt{2}}>4^{2\sqrt{3}}\\ \Rightarrow4^{3\sqrt{2}}>16^{\sqrt{3}}\)

\(c,\left(\sqrt{16}\right)^6=16^3=4^6=4^2\cdot4^4=4^2\cdot16^2\\ \left(\sqrt[3]{60}\right)^6=60^2=4^2\cdot15^2\\ 4^2\cdot16^2>4^2\cdot15^2\Rightarrow\sqrt{16}>\sqrt[3]{60}\Rightarrow0,2^{\sqrt{16}}< 0,2^{\sqrt[3]{60}}\)

`(3\sqrt2)^2.\sqrt(2/(16-6\sqrt7))`

`=18. \sqrt(2/(9-6\sqrt7+7))`

`=18 .\sqrt(2/(3^2-2.3.\sqrt7+\sqrt7^2))`

`=18 .\sqrt(2/((3-\sqrt7)^2))`

`= 18. \sqrt2/(3-\sqrt7)`

`=18 . (\sqrt14+3\sqrt2)/2`

`=9+\sqrt14+27\sqrt2`

\(\left(3\sqrt{2}\right)^2\cdot\sqrt{\dfrac{2}{16-6\sqrt{7}}}\)

\(=18\cdot\dfrac{\sqrt{2}}{3-\sqrt{7}}\)

\(=\dfrac{18\sqrt{2}}{3-\sqrt{7}}=9\sqrt{14}+27\sqrt{2}\)