Question

Câu 5: Cho biểu thức: \(A=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)

a.Rút gọn biểu thức A

b.Tìm giá trị A biết \(a=4+2\sqrt{3}\)

c.Tìm a để \(A< 0\)

Answer 1

a) Ta có: \(A=\left(\frac{\sqrt{a}}{\sqrt{a}-1}-\frac{1}{a-\sqrt{a}}\right):\left(\frac{1}{\sqrt{a}+1}+\frac{2}{a-1}\right)\)

\(=\left(\frac{a}{\sqrt{a}\left(\sqrt{a}-1\right)}-\frac{1}{\sqrt{a}\left(\sqrt{a}-1\right)}\right):\left(\frac{\sqrt{a}-1}{\left(\sqrt{a}+1\right)\left(\sqrt{a}-1\right)}+\frac{2}{\left(\sqrt{a}-1\right)\left(\sqrt{a}+1\right)}\right)\)

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

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

\(=\frac{a-1}{\sqrt{a}}\)