a) ĐKXĐ: \(a\ge0;a\ne1\)
\(A=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(\Leftrightarrow A=\left(1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right)\left(1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right)\)
\(\Leftrightarrow A=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(\Leftrightarrow A=1-a\)
b) \(A=-2010\)
\(\Leftrightarrow1-a=-2010\)
\(\Leftrightarrow a=2011\) (thỏa mãn)
Kết luận ...
\(a.\)
\(A=\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{\sqrt{a}-1}\right)\)
\(A=\left[1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right]\left[1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right]\)
\(A=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(A=1-a\)
\(b.\)
Để \(N=-2010\Rightarrow1-a=-2010\Rightarrow a=2011\)
