Question

Cho A= (\(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\)).(\(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\) -1)

Với 1 khác a lớn bằng 0. Tìm a sao cho A = -a²

Answer 1

A = \(\left(\dfrac{a+\sqrt{a}}{\sqrt{a}+1}+1\right).\left(\dfrac{a-\sqrt{a}}{\sqrt{a}-1}-1\right)\) điều kiện \(a>0vàa\ne1\)

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

A = \(\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)\) = \(a-1\)

ta có : A = \(-a^2\Leftrightarrow a-1=-a^2\Leftrightarrow a^2+a-1=0\)

\(\Delta\) = \(1^2-4.1.\left(-1\right)=1+4=5\)

\(a_1=\dfrac{-1+\sqrt{5}}{2}\) (tmđk)

\(a_2=\dfrac{-1-\sqrt{5}}{2}\) (loại)

\(vậya=\dfrac{-1+\sqrt{5}}{2}thìA=-a^2\)