Question

[Nguyễn Ngọc Nhã Hân]

Thực hiện phép tính:

a) \(\left(2\sqrt{2}-\sqrt{3}\right)^2\)

b) \(\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)\)

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

d) \(\left(\sqrt{\sqrt{11}+\sqrt{7}}-\sqrt{\sqrt{11}-\sqrt{7}}\right)^2\)

e) \(\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}\)

f) \(\sqrt{21-12\sqrt{3}}-\sqrt{3}\)

g) (\(\sqrt{6}+\sqrt{2}\))(\(\sqrt{3}-2\))\(\sqrt{\sqrt{3}+2}\)

h) \(\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}\)

Mọi người làm giúp em gấp với !!!!!!!!!!!!!!!!!

Answer 1

[Phùng Khánh Linh]

\(a.\left(2\sqrt{2}-\sqrt{3}\right)^2=8-4\sqrt{6}+3=11-4\sqrt{6}\)

\(b.\left(1+\sqrt{3}-\sqrt{2}\right)\left(1+\sqrt{3}+\sqrt{2}\right)=\left(1+\sqrt{3}\right)^2-2=4+2\sqrt{3}-2=2+2\sqrt{3}\) \(c.\left(\sqrt{3-\sqrt{5}}+\sqrt{3+\sqrt{5}}\right)^2=3-\sqrt{5}+3+\sqrt{5}+2\sqrt{9-5}=6+4=10\) \(d.\left(\sqrt{\sqrt{11}+\sqrt{7}}-\sqrt{\sqrt{11}-\sqrt{7}}\right)^2=\sqrt{11}+\sqrt{7}+\sqrt{11}-\sqrt{7}-2\sqrt{11-7}=2\sqrt{11}-4\) \(e.\sqrt{2+\sqrt{3}}-\sqrt{2-\sqrt{3}}=\dfrac{\sqrt{3+2\sqrt{3}+1}-\sqrt{3-2\sqrt{3}+1}}{\sqrt{2}}=\dfrac{\sqrt{3}+1-\sqrt{3}+1}{\sqrt{2}}=\sqrt{2}\) \(f.\sqrt{21-12\sqrt{3}}-\sqrt{3}=\sqrt{12-2.2\sqrt{3}.3+9}-\sqrt{3}=2\sqrt{3}-3-\sqrt{3}=\sqrt{3}-3\)

\(g.\left(\sqrt{6}+\sqrt{2}\right)\left(\sqrt{3}-2\right)\sqrt{\sqrt{3}+2}=\left(\sqrt{3}+1\right)\left(\sqrt{3}-2\right)\sqrt{3+2\sqrt{3}+1}=\left(\sqrt{3}+1\right)^2\left(\sqrt{3}-2\right)=\left(4+2\sqrt{3}\right)\left(\sqrt{3}-2\right)=2\left(2+\sqrt{3}\right)\left(\sqrt{3}-2\right)=2\left(3-4\right)=-2\)

\(h.\sqrt{6-2\sqrt{\sqrt{2}+\sqrt{12}+\sqrt{18-\sqrt{128}}}}=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+\sqrt{16-2.4\sqrt{2}+2}}}=\sqrt{6-2\sqrt{\sqrt{2}+2\sqrt{3}+4-\sqrt{2}}}=\sqrt{6-2\sqrt{3+2\sqrt{3}+1}}=\sqrt{6-2\left(\sqrt{3}+1\right)}=\sqrt{3-2\sqrt{3}+1}=\sqrt{3}-1\)