Question

[Trần Linh Nhi]

Cho biểu thức D= \(\left[\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right]\)\(:\left[1+\frac{a+b+2ab}{1-ab}\right]\)

a) Tính giá trị D với a= \(\frac{2}{2+\sqrt{3}}\)

b) Tính giá trị lớn nhất của D

Answer 1

[Hoàng Lê Bảo Ngọc]

 Trước hết ta rút gọn D :

 \(D=\left(\frac{\sqrt{a}+\sqrt{b}}{1-\sqrt{ab}}+\frac{\sqrt{a}-\sqrt{b}}{1+\sqrt{ab}}\right):\left(1+\frac{a+b+2ab}{1-ab}\right)\)(ĐKXĐ : \(a\ne0,b\ne0,ab\ne1\))

\(=\frac{\left(\sqrt{a}+\sqrt{b}\right)\left(1+\sqrt{ab}\right)+\left(\sqrt{a}-\sqrt{b}\right)\left(1-\sqrt{ab}\right)}{\left(1-\sqrt{ab}\right)\left(1+\sqrt{ab}\right)}:\frac{1+a+b+ab}{1-ab}\)

\(=\frac{2\sqrt{a}\left(b+1\right)}{1-ab}.\frac{1-ab}{\left(a+1\right)\left(b+1\right)}=\frac{2\sqrt{a}}{a+1}\)

a) Với \(a=\frac{2}{2+\sqrt{3}}=\frac{2\left(2-\sqrt{3}\right)}{4-3}=4-2\sqrt{3}=\left(\sqrt{3}-1\right)^2\)

\(\Rightarrow D=\frac{2\sqrt{\left(\sqrt{3}-1\right)^2}}{4-2\sqrt{3}+1}=\frac{2\left(\sqrt{3}-1\right)}{5-2\sqrt{3}}\)

b) Ta có : \(\left(\sqrt{a}-1\right)^2\ge0\Leftrightarrow a+1\ge2\sqrt{a}\Leftrightarrow\frac{2\sqrt{a}}{a+1}\le1\)

Suy ra Max D = 1 <=> a = 1