1, A=\(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{a\sqrt{a}+\sqrt{a}+a+1}\right)\)
ĐKXĐ: a≥0
A=\(\left(1-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{1}{\sqrt{a}+1}-\dfrac{2\sqrt{a}}{\sqrt{a}\left(a+1\right)+1\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1}{a+1}-\dfrac{2\sqrt{a}}{a+1}\right):\left(\dfrac{a+1}{\left(\sqrt{a}+1\right)\left(a+1\right)}-\dfrac{2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1-2\sqrt{a}}{a+1}\right):\left(\dfrac{a+1-2\sqrt{a}}{\left(\sqrt{a}+1\right)\left(a+1\right)}\right)\)
A=\(\left(\dfrac{a+1-2\sqrt{a}}{a+1}\right).\left(\dfrac{\left(a+1\right)\left(\sqrt{a}+1\right)}{a+1-2\sqrt{a}}\right)\)
A=\(\sqrt{a}+1\)
Vậy A=\(\sqrt{a}+1\)
2, a=1996-2\(\sqrt{1995}\)
a=\(1995-2\sqrt{1995}+1\)
a=\(\left(\sqrt{1995}-1\right)^2\) (TMĐKXĐ)
thay a=\(\left(\sqrt{1995}-1\right)^2\) vào A ta có:
A=\(\sqrt{\left(\sqrt{1995}-1\right)^2}+1\)
A=\(\sqrt{1995}\)
Vậy a=1996-2\(\sqrt{1995}\) thì A=\(\sqrt{1995}\)
