\(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{3}\right)^2+2\sqrt{3}\sqrt{2}+\left(\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{3}+\sqrt{2}\right)^2}+\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{3}+\sqrt{2}+\sqrt{5}-\sqrt{3}=\sqrt{2}+\sqrt{5}\)

\(\sqrt{4+2\sqrt{3}}+\sqrt{4-2\sqrt{3}}-\dfrac{5}{\sqrt{3}-2\sqrt{2}}-\dfrac{5}{\sqrt{3}+\sqrt{8}}=\sqrt{\sqrt{3}^2+2\sqrt{3}.1+1^2}+\sqrt{\sqrt{3}^2-2\sqrt{3}.1+1^2}-\dfrac{5\left(\sqrt{3}+2\sqrt{2}\right)}{\left(\sqrt{3}-2\sqrt{2}\right)\left(\sqrt{3}+2\sqrt{2}\right)}-\dfrac{5\left(\sqrt{3}-2\sqrt{2}\right)}{\left(\sqrt{3}+2\sqrt{2}\right)\left(\sqrt{3}-2\sqrt{2}\right)}=\sqrt{\left(\sqrt{3}+1\right)^2}+\sqrt{\left(\sqrt{3}-1\right)^2}-\dfrac{5\sqrt{3}+10\sqrt{2}}{9-8}-\dfrac{5\sqrt{3}-10\sqrt{2}}{9-8}=\sqrt{3}+1+\sqrt{3}-1-5\sqrt{3}-10\sqrt{2}-5\sqrt{3}+10\sqrt{2}=-8\sqrt{3}\)\(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}\right)^2+2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}\right)^2-2\sqrt{5}\sqrt{3}+\left(\sqrt{3}\right)^2}=\sqrt{\left(\sqrt{5}+\sqrt{3}\right)^2}-\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}=2\sqrt{3}\)

Bài 1:

Ta có: \(A=\sqrt{8+2\sqrt{15}}+\sqrt{8-2\sqrt{15}}-2\sqrt{6-2\sqrt{5}}\)

\(=\sqrt{5+2\cdot\sqrt{5}\cdot\sqrt{3}+3}+\sqrt{5-2\cdot\sqrt{5}\cdot\sqrt{3}+3}-2\cdot\sqrt{5-2\cdot\sqrt{5}\cdot1+1}\)

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

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

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

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

=2

Vậy: A=2

Bài 2: Sửa đề: Chứng minh \(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right)\cdot\frac{1}{\sqrt{6}}=\frac{-7+\sqrt{3}}{6}\)

Ta có: \(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\left(\frac{9\sqrt{2}}{3\left(\sqrt{27}-3\right)}-\frac{\sqrt{150}\left(\sqrt{27}-3\right)}{3\cdot\left(\sqrt{27}-3\right)}\right)\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{9\sqrt{2}-45\sqrt{2}+3\sqrt{150}}{9\left(\sqrt{3}-1\right)}\cdot\frac{1}{\sqrt{6}}\)

\(=\frac{-36\sqrt{2}+3\sqrt{150}}{9\sqrt{6}\cdot\left(\sqrt{3}-1\right)}\)

\(=\frac{\sqrt{54}\cdot\left(5-4\sqrt{3}\right)}{\sqrt{486}\cdot\left(\sqrt{3}-1\right)}\)

\(=\frac{5-4\sqrt{3}}{3\sqrt{3}-3}\)

\(=\frac{-7+\sqrt{3}}{6}\)(đpcm)

1/ \(A=\sqrt{8-2\sqrt{15}}=\sqrt{\left(\sqrt{5}-\sqrt{3}\right)^2}=\left|\sqrt{5}-\sqrt{3}\right|=\sqrt{5}-\sqrt{3}\) (Vì \(\sqrt{5}-\sqrt{3}>0\))

\(B=\sqrt{6+2\sqrt{5}}-\sqrt{13}+\sqrt{48}=\sqrt{\left(\sqrt{5}+1\right)^2}-\sqrt{13}+4\sqrt{3}=\left|\sqrt{5}+1\right|-\sqrt{13}+4\sqrt{3}=\sqrt{5}+1+\sqrt{13}+4\sqrt{5}\)

2/Ta có :

\(\left(\frac{3\sqrt{2}}{\sqrt{27}-3}-\frac{\sqrt{150}}{3}\right).\frac{1}{\sqrt{6}}\)

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

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

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

\(=\frac{-12\sqrt{2}+5\sqrt{6}}{3\left(\sqrt{3}-1\right)}.\frac{1}{\sqrt{6}}\)

\(=\frac{-7+\sqrt{3}}{6}\)

Vậy...

\(\left(3\sqrt{12}-4\sqrt{3}+\sqrt{15}\right)\cdot\sqrt{3}-2\sqrt{5}\)

\(=\left(6\sqrt{3}-4\sqrt{3}+\sqrt{15}\right)\cdot\sqrt{3}-2\sqrt{5}\)

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

(3\(\sqrt{12}\)-4\(\sqrt{3}\)+\(\sqrt{15}\)).\(\sqrt{3}\)-2\(\sqrt{5}\)

=\(\left(6\sqrt{3}-4\sqrt{3}+\sqrt{15}\right).\sqrt{3}-2\sqrt{5}\)

=\(\left(2\sqrt{3}+\sqrt{15}\right).\sqrt{3}-2\sqrt{5}\)

=\(6+\sqrt{45}-2\sqrt{5}\)

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

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

Đặt \(A=\frac{T}{M}\), ta có T>0 => \(T=\sqrt{T^2}\). Xét

\(T^2=\left(\sqrt[4]{8}+\sqrt{\sqrt{2}-1}\right)-2\sqrt{\left(\sqrt[4]{8}+\sqrt{\sqrt{2}-1}\right)}+\left(\sqrt[4]{8}-\sqrt{\sqrt{2}-1}\right)\)

\(=2\sqrt[4]{8}-2\sqrt{\sqrt{8}-\left(\sqrt{2}-1\right)}\)

\(=2\sqrt[4]{8}-2\sqrt{\sqrt{2}+1}\)

\(=2\left(\sqrt[4]{8}-\sqrt{\sqrt{2}+1}\right)\)

\(\Rightarrow T=\sqrt{2}\cdot\sqrt{\sqrt[4]{8}-2\sqrt{2}+1}\)

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

:) vẫn sắc sảo như mọi khi 

\(A=\dfrac{\sqrt{8}-\sqrt{7}}{8-7}+\sqrt{25\times7}-2\sqrt{2}\\ =\sqrt{8}-\sqrt{7}+5\sqrt{7}-2\sqrt{2}\\ =2\sqrt{2}-\sqrt{7}+5\sqrt{7}-2\sqrt{2}\\ =4\sqrt{7}\)

A=\(\sqrt{5-2\sqrt{3}.\sqrt{5}+3}-\sqrt{5+2\sqrt{5}.\sqrt{3}+3}\)

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

A=\(\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)

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

\(A=\sqrt{8-2\sqrt{15}}-\sqrt{8+2\sqrt{15}}\)

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

\(A=\left|\sqrt{5}-\sqrt{3}\right|-\sqrt{5}-\sqrt{3}\)

\(A=\sqrt{5}-\sqrt{3}-\sqrt{5}-\sqrt{3}\)

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

4: \(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\)

\(=\sqrt{5}+\sqrt{3}-\sqrt{5}+\sqrt{3}\)

\(=2\sqrt{3}\)

4) \(\sqrt{8+2\sqrt{15}}-\sqrt{8-2\sqrt{15}}\)

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

5) \(\sqrt{5+2\sqrt{6}}+\sqrt{8-2\sqrt{15}}\)

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

binh phuong len di ban

\(\frac{5}{\sqrt{2}-7}-\frac{4}{3\sqrt{2}+5}-\frac{7}{4-5\sqrt{2}}\)