Question

[Yuu~chan]

Giúp mình vs ạ. Cảm ơn mọi người nhiều

Answer 1

[An Thy]

1) \(\sqrt{2-\sqrt{3}}\left(\sqrt{5}+\sqrt{2}\right)=\sqrt{\dfrac{4-2\sqrt{3}}{2}}\left(\sqrt{5}+\sqrt{2}\right)\)

\(=\sqrt{\dfrac{\left(\sqrt{3}-1\right)^2}{2}}\left(\sqrt{5}+\sqrt{2}\right)=\dfrac{\sqrt{3}-1}{\sqrt{2}}\left(\sqrt{5}+\sqrt{2}\right)=\dfrac{\sqrt{15}+\sqrt{6}-\sqrt{5}-\sqrt{2}}{\sqrt{2}}\)

theo mình nghĩ thì đề nên là \(\sqrt{2-\sqrt{3}}\left(\sqrt{6}+\sqrt{2}\right)\)

như thế thì sẽ  \(=\dfrac{\sqrt{3}-1}{\sqrt{2}}.\sqrt{2}\left(\sqrt{3}+1\right)=\left(\sqrt{3}-1\right)\left(\sqrt{3}+1\right)=2\) đẹp hơn,đó là mình nghĩ vậy thôi,còn nếu đề bạn đúng thì mình làm ở trên đó

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

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

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

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

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

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

\(\Rightarrow A=\sqrt{5}+1\left(A\ge0\right)\)

4) \(\dfrac{\left(\sqrt{5}+2\right)^2-8\sqrt{5}}{2\sqrt{5}-4}=\dfrac{9+4\sqrt{5}-8\sqrt{5}}{2\left(\sqrt{5}-2\right)}=\dfrac{9-4\sqrt{5}}{2\left(\sqrt{5}-2\right)}\)

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

Answer 2

[Yuu~chan]

Cảm ơn ban ạ