a) Áp dụng công thức nhị thức Newton, ta có

          \(\begin{array}{l}{\left( {2 + \sqrt 2 } \right)^4} = {2^4} + {4.2^3}.\left( {\sqrt 2 } \right) + {6.2^2}.{\left( {\sqrt 2 } \right)^2} + 4.2.{\left( {\sqrt 2 } \right)^3} + {\left( {\sqrt 2 } \right)^4}\\ = \left[ {{2^4} + {{6.2}^2}.{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^4}} \right] + \left[ {{{4.2}^3}.\left( {\sqrt 2 } \right) + 4.2.{{\left( {\sqrt 2 } \right)}^3}} \right]\\ = 68 + 48\sqrt 2 \end{array}\)

b) Áp dụng công thức nhị thức Newton, ta có

          \({\left( {2 + \sqrt 2 } \right)^4} = {2^4} + {4.2^3}.\left( {\sqrt 2 } \right) + {6.2^2}.{\left( {\sqrt 2 } \right)^2} + 4.2.{\left( {\sqrt 2 } \right)^3} + {\left( {\sqrt 2 } \right)^4}\)

          \({\left( {2 - \sqrt 2 } \right)^4} = \left( {2 +(- \sqrt 2 )} \right)^4= {2^4} + {4.2^3}.\left( { - \sqrt 2 } \right) + {6.2^2}.{\left( { - \sqrt 2 } \right)^2} + 4.2.{\left( { - \sqrt 2 } \right)^3} + {\left( { - \sqrt 2 } \right)^4}\)

Từ đó,

          \(\begin{array}{l}{\left( {2 + \sqrt 2 } \right)^4} + {\left( {2 - \sqrt 2 } \right)^4} = 2\left[ {{2^4} + {{6.2}^2}.{{\left( {\sqrt 2 } \right)}^2} + {{\left( {\sqrt 2 } \right)}^4}} \right]\\ = 2\left( {16 + 48 + 4} \right) = 136\end{array}\)

c) Áp dụng công thức nhị thức Newton, ta có

          \(\begin{array}{l}{\left( {1 - \sqrt 3 } \right)^5} = \left( {1 +(- \sqrt 3 )} \right)^5=  1 + 5.\left( { - \sqrt 3 } \right) + 10.{\left( { - \sqrt 3 } \right)^2} + 10.{\left( { - \sqrt 3 } \right)^3} + 5.{\left( { - \sqrt 3 } \right)^4} + 1.{\left( { - \sqrt 3 } \right)^5}\\ = \left[ {1 + 10.{{\left( { - \sqrt 3 } \right)}^2} + 5.{{\left( { - \sqrt 3 } \right)}^4}} \right] + \left[ {5.\left( { - \sqrt 3 } \right) + 10.{{\left( { - \sqrt 3 } \right)}^3} + 1.{{\left( { - \sqrt 3 } \right)}^5}} \right]\\ = 76 - 44\sqrt 3 \end{array}\)

\(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)

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

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

\(=\frac{\sqrt{3}.2}{\sqrt{3}}\)

\(=2\)

\(\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}-1}-\frac{\sqrt{3}}{\sqrt{\sqrt{3}+1}+1}\)

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

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

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

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

= 2

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

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

\(\Leftrightarrow A^2=8+2\sqrt{16-10-2\sqrt{5}}\\ \Leftrightarrow A^2=8+2\sqrt{6-2\sqrt{5}}\\ \Leftrightarrow A^2=8+2\left(\sqrt{5}-1\right)\\ \Leftrightarrow A^2=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\\ \Leftrightarrow A=\sqrt{5}+1\)

Vậy \(\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}=\sqrt{5}+1\)

#)Giải :

Bình phương hai vế, ta được : 

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

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

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

Do \(B>0\)nên \(B=\sqrt{8+2\left(\sqrt{5}-1\right)}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)

           #~Will~be~Pens~#

Bình phương hai vế, ta được:
B2=8+2√(4+√10+2√5)(4−√10+2√5)=8+2√(16−(10+2√5))B2=8+2(4+10+25)(4−10+25)=8+2(16−(10+25))
B2=8+2√6−2√5=8+2√(√5−1)2=8+2(√5−1)B2=8+26−25=8+2(5−1)2=8+2(5−1)
Do B>0B>0 nên B=√8+2(√5−1)=√6+2√5=√5+1B=8+2(5−1)=6+25=5+1

Tk mk nha 

~ Hok tốt ~

Thanks m.n đã tk mk

Bạn tham khảo lời giải tại đây:

Câu hỏi của khanhhuyen6a5 - Toán lớp 9 | Học trực tuyến

Đặt biểu thức trên là \(A\)

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

\(=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}=8+2\sqrt{6-2\sqrt{5}}\)

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

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

\(\Rightarrow A=\sqrt{6+2\sqrt{5}}=\sqrt{\left(\sqrt{5}+1\right)^2}=\sqrt{5}+1\)

cho hỏi sao ra được kết quả như vậy giải thích dùm đi

Gọi biểu thức là P:

Bình phương hai vế, ta có: 

\(P^2=8+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}=8+2\sqrt{16-\left(10+2\sqrt{5}\right)}\)

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

\(\text{Do }P>0\text{ nên }P=\sqrt{8+2\left(\sqrt{5}-1\right)}=\sqrt{6+2\sqrt{5}}=\sqrt{5}+1\)

Đặt cái đấy là A

A² = 8 + \(2\sqrt{6-2\sqrt{5}}\)

= 8 + \(2\sqrt{5}-2\)

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

=> A = \(1+\sqrt{5}\)

bằng 4,877630889.10^-4

Rút gọn mà . Ai nói dùng máy

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

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

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

\(=8+2\sqrt{6-2\sqrt{5}}\)

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

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

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

\(\Rightarrow A=\sqrt{6+2\sqrt{6}}\)