cho A= (1/1-căn a+1/1+căn a):(1/1-căn a-1/1+căn a)+1/1-căn a a.rút gọn A b.Tính giá trị của A khi a =7+4 căn 3

Đáp án:

$a)A=\dfrac{1}{\sqrt{a}(1-\sqrt{a})}\\ b)A=\dfrac{1}{1+\sqrt{3}}$

Giải thích các bước giải:

$A=\left(\dfrac{1}{1-\sqrt{a}}+\dfrac{1}{1+\sqrt{a}}\right):\left(\dfrac{1}{1-\sqrt{a}}-\dfrac{1}{1+\sqrt{a}}\right)+\dfrac{1}{1-\sqrt{a}}\\ =\left(\dfrac{1+\sqrt{a}}{(1-\sqrt{a})(1+\sqrt{a})}+\dfrac{1-\sqrt{a}}{(1-\sqrt{a})(1+\sqrt{a})}\right):\left(\dfrac{1+\sqrt{a}}{(1-\sqrt{a})(1+\sqrt{a})}-\dfrac{1-\sqrt{a}}{(1-\sqrt{a})(1+\sqrt{a})}\right)+\dfrac{1}{1-\sqrt{a}}\\ =\dfrac{2}{(1-\sqrt{a})(1+\sqrt{a})}:\dfrac{2\sqrt{a}}{(1-\sqrt{a})(1+\sqrt{a})}+\dfrac{1}{1-\sqrt{a}}\\ =\dfrac{1}{\sqrt{a}}+\dfrac{1}{1-\sqrt{a}}\\ =\dfrac{1-\sqrt{a}}{\sqrt{a}(1-\sqrt{a})}+\dfrac{\sqrt{a}}{\sqrt{a}(1-\sqrt{a})}\\ =\dfrac{1}{\sqrt{a}(1-\sqrt{a})}\\ b)7+4\sqrt{3}\\ =4+4\sqrt{3}+3\\ =2^2+2.2.\sqrt{3}+\left(\sqrt{3}\right)^2\\ =\left(2+\sqrt{3}\right)^2\\ A=\dfrac{1}{\sqrt{\left(2+\sqrt{3}\right)^2}\left(1-\sqrt{\left(2+\sqrt{3}\right)^2}\right)}\\ =\dfrac{1}{\left(2+\sqrt{3}\right)\left(1-2+\sqrt{3}\right)}\\ =\dfrac{1}{\left(2+\sqrt{3}\right)\left(\sqrt{3}-1\right)}\\=\dfrac{1}{1+\sqrt{3}}$