Question

Cho A=\(\dfrac{2a^2+4}{1+a^3}-\dfrac{1}{1+\sqrt{a}}-\dfrac{1}{1-\sqrt{a}}\)

a)Rút gọn A

b)Tìm GTLN của A

Answer 1

điều kiện xác định : \(a\ge0;a\ne1\)

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

\(\Leftrightarrow A=\dfrac{2a^2+4}{\left(a+1\right)\left(a^2-a+1\right)}+\dfrac{2}{a-1}=\dfrac{\left(2a^2+4\right)\left(a-1\right)+2\left(a^3+1\right)}{\left(a^2-1\right)\left(a^2-a+1\right)}\)

\(\Leftrightarrow A=\dfrac{4a^3-2a^2+4a-2}{\left(a^2-1\right)\left(a^2-a+1\right)}=\dfrac{\left(a^2+1\right)\left(4a-2\right)}{\left(a^2-1\right)\left(a^2-a+1\right)}\)

tới đây mk nghỉ đề sai rồi bn à