Question

Cho \(C=\left(\dfrac{1+a\sqrt{a}}{1+\sqrt{a}}-\sqrt{a}\right).\dfrac{1+2\sqrt{a}+a}{\left(1-a\right)^2}+\sqrt{a}\)

với a > 0 , a ≠ 1

a. Rút gọn C

b. Tìm a để C = 3

Answer 1

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

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

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

Để \(C=3\Rightarrow\sqrt{a}+1=3\Rightarrow\sqrt{a}=2\Rightarrow a=4\)