Câu hỏi của Trân Vũ Mai Ngọc - Toán lớp 9 - Học toán với OnlineMath

Lời giải:
Gọi biểu thức trên là $A$

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

\(=8+2\sqrt{4^2-(10+2\sqrt{5})}=8+2\sqrt{6-2\sqrt{5}}\)

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

$\Rightarrow A=\sqrt{5}+1$ (do $A>0$)

\(Xét-biểu-thức:=>T=\sqrt{4+\sqrt{10+2\sqrt{5}}}+\sqrt{4-\sqrt{10+2\sqrt{5}}}.\\ \)
Bình phương T thì được điều bất ngờ =))))))))))))

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

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

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

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

=\(8+2.\sqrt{5-2\sqrt{5}+1}\)

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

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

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

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

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

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

1.\(\left(\sqrt{2}+1\right)^3-\left(\sqrt{2}-1\right)^3=2\sqrt{2}+6+3\sqrt{2}+1-\left(2\sqrt{2}-6+3\sqrt{2}-1\right)=14\)

2.\(\sqrt{4-\sqrt{15}}+\sqrt{4+\sqrt{15}}-2\sqrt{3-\sqrt{5}}\)

\(=\sqrt{\dfrac{1}{2}\left(8-2\sqrt{3.}\sqrt{5}\right)}+\sqrt{\dfrac{1}{2}\left(8+2.\sqrt{3}.\sqrt{5}\right)}-\sqrt{2}\sqrt{6-2\sqrt{5}}\)

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

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

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

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

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

\(=\sqrt{20}+\dfrac{8\left(1+\sqrt{5}\right)}{-4}=2\sqrt{5}-2\left(1+\sqrt{5}\right)=-2\)

4.\(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

\(=\sqrt{\dfrac{4-2\sqrt{3}}{4+2\sqrt{3}}}+\sqrt{\dfrac{4+2\sqrt{3}}{4-2\sqrt{3}}}\)\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{\left(\sqrt{3}+1\right)^2}}+\sqrt{\dfrac{\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}-1\right)^2}}\)

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

\(=\dfrac{\left(\sqrt{3}-1\right)^2+\left(\sqrt{3}+1\right)^2}{\left(\sqrt{3}+1\right)\left(\sqrt{3}-1\right)}=\dfrac{8}{3-1}=4\)

3: Ta có: \(\dfrac{10+2\sqrt{10}}{\sqrt{5}+\sqrt{2}}+\dfrac{8}{1-\sqrt{5}}\)

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

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

=-2

4) Ta có: \(\sqrt{\dfrac{2-\sqrt{3}}{2+\sqrt{3}}}+\sqrt{\dfrac{2+\sqrt{3}}{2-\sqrt{3}}}\)

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

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

=4

khó quá

\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{16-10-2\sqrt{5}}\)

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

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

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

...

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

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

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

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

=>\(\sqrt{5}+1\)

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

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

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

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

\(\Rightarrow A=\sqrt{5}+1\)

Đặt biểu thứa là A

Bình phương 2 vế ta dc:

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

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

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

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

Do A>0 nên :

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

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