Question

[An Nguyễn Thiện]

Bài 1: Rút gọn

a)\(\sqrt{4+\sqrt{10+2\sqrt{5}}}\)+\(\sqrt{4-\sqrt{10+2\sqrt{5}}}\) ,

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

c)A=\(\sqrt{4+\sqrt{5\sqrt{3}+5\sqrt{48-10\sqrt{7+4\sqrt{3}}}}}\)

d)B=\(\sqrt{6+2\sqrt{2}\sqrt{3-\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-8\sqrt{2}}}}}\)

e)C=\(\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\)

f)D=\(\dfrac{\left(5+2\sqrt{6}\right)\left(49-20\sqrt{6}\right)\left(5-2\sqrt{6}\right)\sqrt{5-2\sqrt{6}}}{9\sqrt{3}-11\sqrt{2}}\)

Mình sửa lại để m.n dễ nhìn hơn!

Answer 1

[Mysterious Person]

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

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

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

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

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

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

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

e) \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{29-12\sqrt{5}}}}\) \(C=\sqrt{\sqrt{5}-\sqrt{3-\sqrt{\left(2\sqrt{5}-3\right)^2}}}\)

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

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

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

câu a ; f chưa nghỉ ra

Answer 2

[An Nguyễn Thiện]

co giup mk nha