Question

[Nguyễn Hoàng trung]

Rút gọn rồi tính các biểu thức sau:

a)\(A=\left(\dfrac{a\sqrt{a}+b\sqrt{b}}{\sqrt{a}+\sqrt{b}}\right)\left(\dfrac{\sqrt{a}+\sqrt{b}}{a-b}\right)^2\) với \(a^2=6-3\sqrt{3};b^2=2+\sqrt{3}\)

b)\(B=\dfrac{\sqrt{2x+2\sqrt{x^2-4}}}{\sqrt{x^2-4}+x+2}\)với \(x=1+\sqrt{5}\)

Answer 1

[Akai Haruma]

Câu a, bạn coi lại đề xem $a^2=6-3\sqrt{3}$ hay $a=6-3\sqrt{3}$???

Answer 2

[Akai Haruma]

b.

\(B=\frac{\sqrt{(x-2)+(x+2)+2\sqrt{(x-2)(x+2)}}}{\sqrt{x^2-4}+x+2}\)

\(=\frac{\sqrt{(\sqrt{x-2}+\sqrt{x+2})^2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x^2-4}+x+2}=\frac{\sqrt{x-2}+\sqrt{x+2}}{\sqrt{x+2}(\sqrt{x-2}+\sqrt{x+2})}=\frac{1}{\sqrt{x+2}}\)

\(=\frac{1}{\sqrt{3+\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{6+2\sqrt{5}}}=\frac{\sqrt{2}}{\sqrt{(\sqrt{5}+1)^2}}=\frac{\sqrt{2}}{\sqrt{5}+1}\)