mình sửa lại đề vì đề sai

\(\sqrt{53+12\sqrt{10}}-\sqrt{47-6\sqrt{10}}=\sqrt{53+2\sqrt{360}}-\sqrt{47-2\sqrt{90}}=\sqrt{45+2\sqrt{45}\sqrt{8}+8}-\sqrt{45-2\sqrt{45}\sqrt{2}+2}=\sqrt{45}+2\sqrt{2}-\sqrt{45}+\sqrt{2}=3\sqrt{2}\)

√ 53 + 12 √ 10 − √ 47 − 6 √ 10 = √ 53 + 2 √ 360 − √ 47 − 2 √ 90 = √ 45 + 2 √ 45 √ 8 + 8 − √ 45 − 2 √ 45 √ 2 + 2 = √ 45 + 2 √ 2 − √ 45 + √ 2 = 3 √ 2

Ta có: \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+\sqrt{2}}\)

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

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

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

\(=\dfrac{\sqrt{2}\left(3\sqrt{6}+3+\sqrt{30}+\sqrt{5}\right)-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{6\sqrt{3}+3\sqrt{2}+2\sqrt{15}+\sqrt{10}-\sqrt{15}-\sqrt{5}}{\sqrt{2}\left(\sqrt{6}+1\right)}\)

\(=\dfrac{6\sqrt{3}+3\sqrt{2}+\sqrt{15}+\sqrt{10}-\sqrt{5}}{ }\)

Đề sai rồi bạn

Ta có: \(\dfrac{2\sqrt{3-\sqrt{5}}\left(3+\sqrt{5}\right)}{\sqrt{10}-\sqrt{2}}-\dfrac{\sqrt{15}+\sqrt{5}}{\sqrt{12}+2}\)

\(=\dfrac{\left(6+2\sqrt{5}\right)\sqrt{6-2\sqrt{5}}}{\sqrt{20}-2}-\dfrac{\sqrt{5}\left(\sqrt{3}+1\right)}{2\left(\sqrt{3}+1\right)}\)

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

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

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

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

a)  \(\sqrt{6+2\sqrt{5}}+\sqrt{6-2\sqrt{5}}\)

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

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

b) \(\sqrt{13+4\sqrt{10}}+\sqrt{13-4\sqrt{10}}\)

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

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

c) \(\sqrt{8\sqrt{3}}-2\sqrt{25\sqrt{12}}+4\sqrt{\sqrt{192}}\)

\(=2\sqrt{2\sqrt{3}}-10\sqrt{2\sqrt{3}}+8\sqrt{2\sqrt{3}}=0\)

giải ra chưa chỉ mình với

Bạn áp dụng hằng đẳng thức (a+b+c)^2= a^2+b^2+c^2+2(ab+ac+bc)

Lời giải:

$(4+\sqrt{15})(\sqrt{10}-\sqrt{6})\sqrt{4-\sqrt{15}}$

$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{8-2\sqrt{15}}$

$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})\sqrt{(\sqrt{5}-\sqrt{3})^2}$

$=(4+\sqrt{15})(\sqrt{5}-\sqrt{3})(\sqrt{5}-\sqrt{3})$

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

$=2(4^2-15)=2$ (đpcm)