Nhớ không nhầm thì gọi là trục căn thức ở mẫu thì phải, cậu dở lại lý thuyết coi nha :v

\(A=\frac{\sqrt{5}+\sqrt{3}}{\sqrt{5}-\sqrt{3}}+\frac{\sqrt{5}-3}{\sqrt{5}+\sqrt{3}}\)

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

\(=\frac{5+2\sqrt{18}+3}{5-3}+\frac{5-2\sqrt{18}+3}{5-3}\)

\(=\frac{8+6\sqrt{2}}{2}+\frac{8-6\sqrt{2}}{2}\)

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

\(=8\)

Vậy...

Có cách giải nhưng t ko chắc đâu nhá;) đã bảo đưa dạng a, b, c rồi mà cứ đưa dạng này-_-

\(VT< \sqrt{2\sqrt{3\sqrt{4\sqrt{5\sqrt{6....}}}}}=x>0\) (vô hạn dấu căn). Ta sẽ chứng minh x < 3

Ta thấy \(x^2=\sqrt{2}.x\Rightarrow x\left(x-\sqrt{2}\right)=0\Rightarrow x=\sqrt{2}< 3\Rightarrow\text{đpcm }\)

\(x^2=2\sqrt{3\sqrt{4\sqrt{5....\sqrt{2000}}}}ma?\)

link tha khảo 

link : https://olm.vn/hoi-dap/detail/69408192260.html

hok tốt

Mấy bài này rất dài , đăng từ từ thôi nhé bạn .

\(1.\dfrac{\sqrt{30}-\sqrt{2}}{\sqrt{8}-\sqrt{15}}-\sqrt{8-\sqrt{49+8\sqrt{3}}}=\dfrac{\sqrt{60}-\sqrt{4}}{\sqrt{16-2\sqrt{15}}}-\sqrt{8-\sqrt{48+2.4\sqrt{3}+1}}=\dfrac{2\left(\sqrt{15}-1\right)}{\sqrt{\left(\sqrt{15}-1\right)^2}}-\sqrt{8-|4\sqrt{3}+1|}=2-\sqrt{4-2.2\sqrt{3}+3}=2-|2-\sqrt{3}|=\sqrt{3}\)

\(2.\dfrac{2+\sqrt{3}}{\sqrt{2}+\sqrt{2+\sqrt{3}}}+\dfrac{2-\sqrt{3}}{\sqrt{2}-\sqrt{2-\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{\sqrt{4}+\sqrt{4+2\sqrt{3}}}+\dfrac{2\sqrt{2}-\sqrt{6}}{\sqrt{4}-\sqrt{4-2\sqrt{3}}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+\sqrt{\left(\sqrt{3}+1\right)^2}}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-\sqrt{\left(\sqrt{3}-1\right)^2}}=\dfrac{2\sqrt{2}+\sqrt{6}}{2+|\sqrt{3}+1|}+\dfrac{2\sqrt{2}-\sqrt{6}}{2-|\sqrt{3}-1|}=\dfrac{2\sqrt{2}-\sqrt{6}}{3-\sqrt{3}}+\dfrac{2\sqrt{2}+\sqrt{6}}{3+\sqrt{3}}=\dfrac{12\sqrt{2}-2\sqrt{18}}{9-3}=\dfrac{12\sqrt{2}-6\sqrt{2}}{6}=\dfrac{6\sqrt{2}}{6}=\sqrt{2}\)

\(3.\dfrac{\sqrt{2}}{2\sqrt{2}+\sqrt{3+\sqrt{5}}}+\dfrac{\sqrt{2}}{2\sqrt{2}-\sqrt{3-\sqrt{5}}}=\dfrac{2}{4+\sqrt{5+2\sqrt{5}+1}}+\dfrac{2}{4-\sqrt{5-2\sqrt{5}+1}}=\dfrac{2}{4+|\sqrt{5}+1|}+\dfrac{2}{4-|\sqrt{5}-1|}=\dfrac{2}{\sqrt{5}+5}+\dfrac{2}{5-\sqrt{5}}=\dfrac{10-2\sqrt{5}+10+2\sqrt{5}}{20}=\dfrac{20}{20}=1\)

`c)root{3}{4}.root{3}{1-sqrt3}.root{6}{(sqrt3+1)^2}`

`=root{3}{4(1-sqrt3)}.root{3}{1+sqrt3}`

`=root{3}{4(1-sqrt3)(1+sqrt3)}`

`=root{3}{4(1-3)}=-2`

`d)2/(root{3}{3}-1)-4/(root{9}-root{3}{3}+1)`

`=(2(root{3}{9}+root{3}{3}+1))/(3-1)-(4(root{3}{3}+1))/(3+1)`

`=root{3}{9}+root{3}{3}+1-root{3}{3}-1`

`=root{3}{9}`

`a)root{3}{8sqrt5-16}.root{3}{8sqrt5+16}`

`=root{3}{(8sqrt5-16)(8sqrt5+16)}`

`=root{3}{320-256}`

`=root{3}{64}=4`

`b)root{3}{7-5sqrt2}-root{6}{8}`

`=root{3}{1-3.sqrt{2}+3.2.1-2sqrt2}-root{6}{(2)^3}`

`=root{3}{(1-sqrt2)^3}-sqrt2`

`=1-sqrt2-sqrt2=1-2sqrt2`

a)\(\sqrt{\frac{3a}{7}}-2\sqrt{\frac{7a}{3}}+\sqrt{21a}\)  =\(\sqrt{\frac{3}{7}.\frac{1}{21}.21a}\)  -  \(2\sqrt{\frac{7}{3}.\frac{1}{21}.21a}\)+  \(\sqrt{21}\)

=\(\sqrt{\frac{1}{49}.21a}\) -  \(2\sqrt{\frac{1}{9}.21a}\)+\(\sqrt{21}\)

=\(\sqrt{\frac{1}{49}}.\sqrt{21a}\)  -   \(2.\sqrt{\frac{1}{9}}.\sqrt{21a}\)+  \(\sqrt{21a}\)

=\(\frac{1}{7}\sqrt{21a}\) - \(\frac{2}{3}\sqrt{21a}\)  +  \(\sqrt{21a}\)

=\(\frac{-10}{21}\sqrt{21a}\)

b)

N=\(\sqrt{\frac{8x}{3}}\) - \(\sqrt{\frac{27x}{2}}\) + \(\sqrt{6x}\)

=\(\sqrt{\frac{8}{3}.\frac{1}{6}.6x}\) - \(\sqrt{\frac{27}{2}.\frac{1}{6}.6x}\)+ \(\sqrt{6x}\)

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

=\(\frac{1}{6}\sqrt{6x}\)

em lớp 8 nene làm theo cách hiểu thôi ạ

c)P=\(2\sqrt{\frac{8y}{5}}\) + \(\sqrt{\frac{45y}{2}}\) -  \(\sqrt{10y}\)

=\(2\sqrt{\frac{8}{5}.\frac{1}{10}.10y}\) + \(\sqrt{\frac{45}{2}.\frac{1}{10}.10y}\) -  \(\sqrt{10y}\)

=\(2\sqrt{\frac{4}{25}.10y}\) + \(\sqrt{\frac{9}{4}.10y}\) - \(\sqrt{10y}\)

=\(2\).\(\sqrt{\frac{4}{25}}\)   \(.\sqrt{10y}\) + \(\sqrt{\frac{9}{4}}.\sqrt{10y}\) - \(\sqrt{10y}\)

=\(\frac{4}{5}\sqrt{10y}\) + \(\frac{3}{2}\sqrt{10y}\) - \(\sqrt{10y}\)

=\(\frac{13}{10}\sqrt{10y}\)

a: \(A=\dfrac{2\sqrt{2}\left(\sqrt{3}+1\right)}{3\cdot\sqrt{2+\sqrt{3}}}=\dfrac{4\left(\sqrt{3}+1\right)}{3\cdot\sqrt{4+2\sqrt{3}}}\)

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

b: \(B=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\left|3\sqrt{5}-3\right|\)

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

a: \(=\dfrac{\sqrt{6-2\sqrt{5}}\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}=\dfrac{\left(\sqrt{5}-1\right)\left(3+\sqrt{5}\right)}{2\left(\sqrt{5}+1\right)}\)

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

b: \(=\sqrt{\sqrt{3}}\left(2\sqrt{2}-2\cdot5\sqrt{2}+4\cdot8\sqrt{2}\right)\)

\(=\sqrt{\sqrt{3}}\cdot24\sqrt{2}\)

d: \(=\dfrac{\sqrt{6-2\sqrt{5}}+\sqrt{6+2\sqrt{5}}}{\sqrt{2}}\)

\(=\dfrac{\sqrt{5}-1+\sqrt{5}+1}{\sqrt{2}}=\dfrac{2\sqrt{5}}{\sqrt{2}}=\sqrt{10}\)

a) \(2\sqrt{50}-3\sqrt{32}-\sqrt{162}+5\sqrt{98}\)

=\(2.5\sqrt{2}-3.4\sqrt{2}-9\sqrt{2}+5.7\sqrt{2}\)

= \(10\sqrt{2}-12\sqrt{2}-9\sqrt{2}+35\sqrt{2}\)

= \(24\sqrt{2}\)

b) \(\sqrt{8+2\sqrt{7}}+\sqrt{11-4\sqrt{7}}\)

= \(\sqrt{7+2\sqrt{7}+1}+\sqrt{7-4\sqrt{7}+4}\)

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

= \(\sqrt{7}+1+\sqrt{7}-2\)

= \(2\sqrt{7}-1\)

c) \(\dfrac{10}{\sqrt{5}}+\dfrac{8}{3+\sqrt{5}}-\dfrac{\sqrt{18}-3\sqrt{5}}{\sqrt{2}-\sqrt{5}}\)

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

= 3

a) \(\Leftrightarrow A=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}=3\sqrt{2}\)

b) \(\Leftrightarrow B=\sqrt{7-2\sqrt{12}}+\sqrt{12+2\sqrt{27}}=\sqrt{\left(2-\sqrt{3}\right)^2}+\sqrt{\left(3+\sqrt{3}\right)^2}=2-\sqrt{3}+3+\sqrt{3}=5\)

c) \(\Leftrightarrow C=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{\left(3+\sqrt{5}\right)\left(3-\sqrt{5}\right)}=\dfrac{6}{4}=\dfrac{3}{2}\)

d) \(\Leftrightarrow D=3-\left(-2\right)-5=0\)