\(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\) với \(a\ge0;a\ne1\)
\(=\left[1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right]\left[1+\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right]\)
\(=\left(1+\sqrt{a}\right)\left(1+\sqrt{a}\right)\)
\(=\left(1+\sqrt{a}\right)^2\)
vì đề bài của bạn ko rõ nên tôi sẽ làm theo 2 trường hợp. bạn có thể tham khảo
TH1 : \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1+\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left[1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right]\left[1-\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right]\)\(=\left(1+\sqrt{a}\right)\left(1-\sqrt{a}\right)\)
\(=1-a\)
TH2: \(\left(1+\dfrac{a+\sqrt{a}}{\sqrt{a}+1}\right)\left(1-\dfrac{a-\sqrt{a}}{1-\sqrt{a}}\right)\)
\(=\left[1+\dfrac{\sqrt{a}\left(\sqrt{a}+1\right)}{\sqrt{a}+1}\right]\left[1+\dfrac{\sqrt{a}\left(\sqrt{a}-1\right)}{\sqrt{a}-1}\right]\)
\(=\left(1+\sqrt{a}\right)\left(1+\sqrt{a}\right)\)
\(=\left(\sqrt{a}+1\right)^2\)
\(=a+2\sqrt{a}+1\)
