Question

B= \(\left(\dfrac{3}{\sqrt{1+a}}+\sqrt{1-a}\right)\)/\(\left(\dfrac{3}{\sqrt{1-a^2}}+1\right)\)

a.rút gọn

b. tính giá trị B khi a=\(\dfrac{\sqrt{3}}{2+\sqrt{3}}\)

c. tìm a để \(\sqrt{B}\)>B

Answer 1

a> B=\(\left(\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}\right)\):\(\left(\dfrac{3+\sqrt{1-a^2}}{\sqrt{1-a^2}}\right)\)

=\(\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}\).\(\dfrac{\sqrt{1-a^2}}{3+\sqrt{1-a^2}}\)

= \(\dfrac{\sqrt{1-a^2}}{\sqrt{1+a}}\)

Answer 2

a: \(B=\dfrac{3+\sqrt{1-a^2}}{\sqrt{1+a}}:\dfrac{3+\sqrt{1-a^2}}{\sqrt{1-a^2}}=\sqrt{\dfrac{1-a^2}{1+a}}=\sqrt{1-a}\)

b: \(a=\dfrac{\sqrt{3}}{2+\sqrt{3}}=2\sqrt{3}-3\)

Khi a=2 căn 3-3 thì \(B=\sqrt{1-2\sqrt{3}+3}=\sqrt{4-2\sqrt{3}}=\sqrt{3}-1\)