\(A=\sqrt{8+2\sqrt{10+2\sqrt{5}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}}\)

\(A^2=8+2\sqrt{10+2\sqrt{5}+8-2\sqrt{10+2\sqrt{5}}+}2\sqrt{8+2\sqrt{10+2\sqrt{5}}}.\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)

\(A^2=16+2\left[64-4\left(10+2\sqrt{5}\right)\right]\)

\(A^2=16+128-8\left(10+2\sqrt{5}\right)\)

\(A^2=144-80-16\sqrt{5}\)

\(A^2=64-16\sqrt{5}\)

\(A^2=8+2\sqrt{10+2\sqrt{5}}+8-2.\sqrt{10+2\sqrt{5}}+2\sqrt{64-4\left(10+2\sqrt{5}\right)}\)

\(=16+2\sqrt{24-8\sqrt{5}}=16+2\sqrt{\left(2\sqrt{5}\right)^2-2.2\sqrt{5}+2^2}\)

\(=16+2\sqrt{\left(2\sqrt{5}-2\right)^2}=16+2\left(2\sqrt{5}-2\right)=12+4\sqrt{5}\)

\(=2+2.\sqrt{2}.\sqrt{10}+10\)

\(=\left(\sqrt{2}+\sqrt{10}\right)^2\)

=> \(A=\sqrt{2}+\sqrt{10}\)

Biến đổi vế trái ta có :

 \(\sqrt{8+2\sqrt{10+2\sqrt{5}}}+\sqrt{8-2\sqrt{10+2\sqrt{5}}}\)

= \(\sqrt{2}\left(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\right)\)

Đặt A  = \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}\)

A^2 = \(4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\)

    =  8 + \(2\sqrt{16-\left(10-2\sqrt{5}\right)}\)

     = \(8+2\sqrt{16-10+2\sqrt{5}}\)

     = \(8+2\sqrt{6+2\sqrt{5}}=8+2\sqrt{\left(\sqrt{5}-1\right)^2}=8+2\sqrt{5}-2=6+2\sqrt{5}\)

=> A = \(\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)

=> \(\sqrt{2}A=\sqrt{2}\left(\sqrt{5}+1\right)=\sqrt{10}+\sqrt{2}=VP\) ( ĐPCM) 

haha        

bn thang tran lm sai bước đưa ra hdt :v đúng là phải 16 - ( 10 + 2can5 )

= 16 - 10 - 2can5

c. Ta có: C+E=\(\sqrt{45+\sqrt{2009}}+\sqrt{45-\sqrt{2009}}=\sqrt{\left(\sqrt{\dfrac{49}{2}}+\sqrt{\dfrac{41}{2}}\right)^2}+\sqrt{\left(\sqrt{\dfrac{49}{2}}-\sqrt{\dfrac{41}{2}}\right)^2}=\dfrac{7}{\sqrt{2}}+\dfrac{\sqrt{41}}{\sqrt{2}}+\dfrac{7}{\sqrt{2}}-\dfrac{\sqrt{41}}{\sqrt{2}}=\dfrac{2.7}{\sqrt{2}}=7\sqrt{2}\)

=> đpcm.

3.

Ta có: \(VT=\)\(8+2\sqrt{10+2\sqrt{5}}+8-2\sqrt{10+2\sqrt{5}}\)

\(=8+8+\left(2\sqrt{10+2\sqrt{5}}-2\sqrt{10+2\sqrt{5}}\right)\)

\(=16\ne VP\)

⇒ Đề sai

1. Ta có: \(\sqrt{4x}\)- 3\(\sqrt{x}\)+2\(\sqrt{15x}\)=18

⇌2\(\sqrt{x}\)-3\(\sqrt{x}\) +2\(\sqrt{15x}\)=18

⇌\(-\sqrt{x}\) +2\(\sqrt{15x}\)-15 = 3

⇌-(\(\sqrt{x}\) -2\(\sqrt{15x}\)+15 )=3

⇌(\(\sqrt{x}\)-\(\sqrt{15}\))=-3 (vô lí)

Vậy không tìm được giá trị x thỏa mãn bài toán

2.Ta có: B=\(\dfrac{1}{\sqrt{11-2\sqrt{30}}}-\dfrac{3}{7-2\sqrt{10}}\)

= \(\dfrac{1}{\sqrt{6-2\sqrt{6.5}+5}}-\dfrac{3}{2-2\sqrt{2.5}+5}\)

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

=\(\dfrac{1}{\sqrt{6}-\sqrt{5}}-\dfrac{3}{\sqrt{3}-\sqrt{2}}\)

hình như đề sai

ghi đề sai rồi

\(a,Sửa:\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\\ =\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}\\ =2\sqrt{5}-2-2\sqrt{5}=-2\\ b,=\dfrac{\sqrt{32}-\sqrt{12}}{\sqrt{18}-\sqrt{48}}-\dfrac{\sqrt{5}+\sqrt{27}}{\sqrt{6}\left(\sqrt{5}+\sqrt{27}\right)}\\ =\dfrac{\sqrt{2}\left(4-\sqrt{6}\right)}{\sqrt{3}\left(\sqrt{6}-4\right)}-\dfrac{1}{\sqrt{6}}=\dfrac{\sqrt{6}}{3}-\dfrac{\sqrt{6}}{6}=\dfrac{2\sqrt{6}-\sqrt{6}}{6}=\dfrac{\sqrt{6}}{6}\)

Cm giúp mình với nhé. Mình cảm ơn ạ!

đề bài là j vậy bn

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

a)  \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

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

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

b) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)

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

\(=2\sqrt{2}+\sqrt{5}+2\sqrt{2}-\sqrt{5}=4\sqrt{2}\)

c) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)

\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0\)

\(\left(\sqrt{8}-5\sqrt{2}+\sqrt{20}\right)\sqrt{5}-\left(3\sqrt{\frac{1}{10}}+10\right)=\left(2\sqrt{2}-5\sqrt{2}+2\sqrt{5}\right)\sqrt{5}-\frac{3\sqrt{10}}{10}-10\)

\(=-3\sqrt{10}+10-\frac{3\sqrt{10}}{10}-10=-3\sqrt{10}-\frac{3\sqrt{10}}{10}=-3\sqrt{10}\left(1+\frac{1}{10}\right)=\frac{-33\sqrt{10}}{10}=-3,3\sqrt{10}\)