Question

\(D=\left[\dfrac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\dfrac{\sqrt{a}+\sqrt{b}}{1+\sqrt{ab}}\right]\div\left[1+\dfrac{a+b+2ab}{1-ab}\right]\)

a)Rút gọn:

b)Tính giá trị của \(a=\dfrac{2}{\sqrt{3}+2}\)

Answer 1

Mạn phép ko chép lại đề , mk làm luôn

a) \(D=\left[\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}+\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{1-ab}\right]:\dfrac{a+b+2ab+1-ab}{1-ab}\)\(D=\dfrac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}+1-\sqrt{ab}\right)}{1-ab}.\dfrac{1-ab}{a+b+ab+1}\)

\(D=\dfrac{2\left(\sqrt{a}+\sqrt{b}\right)}{\left(b+1\right)\left(a+1\right)}\)

Answer 2

D=A/B

a)

B=1+(a+b+2ab)/(1-ab)=(a+b+ab)/(1-ab)

dk: a,b≥0; a.b≠1

1/B=(1-ab)/(a+b+ab)

A=√a+√b)[(1+√ab)+(1-√ab)]/(1-ab)=2(√a+√b)/(1-ab)

D=2(√a+√b)/[(a+1)(b+1)]

b)

a=2/(√3+2)=2(2-√3)/[(2+√3)(2-√3)]=2(2-√3)=(√3-1)^2