\(=\dfrac{5-3\sqrt{5}+10+6\sqrt{5}}{\left(\sqrt{5}-3\right)\left(\sqrt{5}+3\right)}-\dfrac{2\sqrt{10}+2}{\sqrt{3}-\sqrt{2}}\\ =\dfrac{15+3\sqrt{5}}{5-9}-\left(2\sqrt{10}+2\right)\left(\sqrt{3}+\sqrt{2}\right)\\ =-2\sqrt{30}-4\sqrt{5}-2\sqrt{3}-2\sqrt{2}-\dfrac{15+3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-16\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15-3\sqrt{5}}{4}\\ =\dfrac{-8\sqrt{30}-19\sqrt{5}-8\sqrt{3}-8\sqrt{2}-15}{4}\)

Đặt \(\sqrt{10+2\sqrt{5}}=t\)

\(VT=\sqrt{4+t}+\sqrt{4-t}\)

\(\Leftrightarrow VT^2=4+t+2\sqrt{\left(4+t\right)\left(4-t\right)}+4-t\)

\(=8+2\sqrt{16-t^2}=8+2\sqrt{6-2\sqrt{5}}\)

\(\Rightarrow VT=\sqrt{8+2\sqrt{6-2\sqrt{5}}}\) (không chắc nha)

tth làm chưa triệt để

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

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

       \(=\sqrt{8+2\sqrt{5}-2}\)

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

      \(=\sqrt{5}+1\)

ko cần đoạn cuối mk tự lm rồi

lê thị thu huyền bn tham khảo nhá:

Bình phương 2 vế lên, ta được:

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

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

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

P/s: Máy lác dấu ngoặc đơn phải dùng tạm ngoặc kép thông cảm

tính cụm căn của 2 lũy thừa rồi trừ; KQ;- 46 (NHỚ K ĐÚNG CHO MÌNH NHA THANKS!)

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

B² = \(4+\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}+4-\sqrt{10+2\sqrt{5}}\)

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

= \(8+2\sqrt{6-2\sqrt{5}}\)

= \(8+2\sqrt{5-2\sqrt{5}+1}\)

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

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

= \(6+2\sqrt{5}\)

= \(5+2\sqrt{5}+1\)

= \(\left(\sqrt{5}+1\right)^2\)

=> B = \(\sqrt{5}+1\)

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

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

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

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

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

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

\(=\left(\sqrt{5}+1\right)^2\Rightarrow A=\sqrt{5}+1\left(A>0\right)\)

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

\(A^2=4+\sqrt{10+2\sqrt{5}}+4-\sqrt{10+2\sqrt{5}}+2\sqrt{\left(4+\sqrt{10+2\sqrt{5}}\right)\left(4-\sqrt{10+2\sqrt{5}}\right)}\\ A^2=8+2\sqrt{16-10-2\sqrt{5}}=8+2\sqrt{6-2\sqrt{5}}\\ A^2=8+2\left(\sqrt{5}-1\right)=6+2\sqrt{5}=\left(\sqrt{5}+1\right)^2\\ A=\sqrt{5}+1\)

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

\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)

\(=3\sqrt{2}+10\sqrt{2}-10\sqrt{2}\)

\(=3\sqrt{2}\)

\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)

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

\(=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}\)

\(=\dfrac{3}{2}\)

\(A=3\sqrt{2}+5\sqrt{8}-2\sqrt{50}\)

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

\(B=\dfrac{1}{3+\sqrt{5}}+\dfrac{1}{3-\sqrt{5}}\)

\(B=\dfrac{3-\sqrt{5}+3+\sqrt{5}}{9-5}=\dfrac{6}{4}=\dfrac{3}{2}\)

\(C=\sqrt{7-4\sqrt{3}}+\sqrt{12+6\sqrt{3}}\)

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

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

gọi biểu thức đó là A , ta có :

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

      \(=8+2\sqrt{5}-2\)

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

       \(\Rightarrow A=1+\sqrt{5}\)

tk mình nhoa bạn

mk nghi  tat ca la 10+2\(\sqrt{5}\)