\(\sqrt{8-4\sqrt{3}}-\sqrt{8+4\sqrt{3}}=\sqrt{5-4\sqrt{3}+3}-\sqrt{5+4\sqrt{3}+3}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}=-2\sqrt{3}\)

\(\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8+10\sqrt{7-4\sqrt{3}}}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8+10\sqrt{4-4\sqrt{3}+3}}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8+10\sqrt{\left(2-\sqrt{3}\right)^2}}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8+10\left(2-\sqrt{3}\right)}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{8}+20-10\sqrt{3}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{28-10\sqrt{3}}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{25-10\sqrt{3}}+3}}=\sqrt{9-\sqrt{5\sqrt{3}+5\sqrt{\left(5-\sqrt{3}\right)^2}}}=\sqrt{9-\sqrt{5\sqrt{3}+5\left(5-\sqrt{3}\right)}}=\sqrt{9-\sqrt{5\sqrt{3}+25-5\sqrt{3}}}=\sqrt{9-\sqrt{25}}=\sqrt{9-5}=\sqrt{4}=2\)

có bạn nào giải nốt câu c giùm k nghĩ mãi k raa

Cm giúp mình với nhé. Mình cảm ơn ạ!

đề bài là j vậy bn

Bài 2: 

a) \(\dfrac{2+\sqrt{2}}{\sqrt{2}+1}=\dfrac{\sqrt{2}\left(\sqrt{2}+1\right)}{\sqrt{2}+1}=\sqrt{2}\)

b) \(\dfrac{a-\sqrt{a}}{1-\sqrt{a}}=\dfrac{-\sqrt{a}\left(1-\sqrt{a}\right)}{1-\sqrt{a}}=-\sqrt{a}\)

c) \(\dfrac{3+\sqrt{3}}{3-\sqrt{3}}+\dfrac{3-\sqrt{3}}{3+\sqrt{3}}\)

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

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

Bài 1: 

a) Đúng

b) Sai vì \(\dfrac{2\sqrt{2}+2}{5\sqrt{2}}=\dfrac{\sqrt{2}\left(2+\sqrt{2}\right)}{5\sqrt{2}}=\dfrac{2+\sqrt{2}}{5}\)

c) Sai vì \(\dfrac{2}{\sqrt{3}-1}=\sqrt{3}+1\)

e) Đúng

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

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

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

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

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

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

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

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

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

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

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

=> \(A=\sqrt{2}+\sqrt{10}\)

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

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

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

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

\(=16-2\sqrt{24-8\sqrt{5}}\)

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

\(=20-4\sqrt{5}\)

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

\(\Rightarrow B>0\)

\(\Rightarrow B=\sqrt{20-4\sqrt{5}}=2\sqrt{5-\sqrt{5}}\)

\(\Rightarrow A=B-\sqrt{2}-\sqrt{10}=2\sqrt{5-\sqrt{5}}-\sqrt{2}-\sqrt{10}=2\)

Câu hỏi của Nguyen Phuc Duy - Toán lớp 9 - Học toán với OnlineMath

Bạn tham khảo link này!

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

đề là rút gọn các biểu thức sau

nhờ mọi người giải giúp mình. cảm ơn mn nhìu

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

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

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

b: \(=\dfrac{2\sqrt{5}\left(\sqrt{5}+\sqrt{2}\right)}{\sqrt{5}+\sqrt{2}}-2-2\sqrt{5}\)

=2căn 5-2-2căn 5

=-2

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

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

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