Question

[Thao Nguyen]

\(A=\left(\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(\sqrt{a}-1\right)}-\frac{a+\sqrt{a}}{a-1}\right):\left(\frac{1}{\sqrt{a}+1}-\frac{1}{\sqrt{a}-1}\right)\)

1. rút gọn

2. tìm a để \(\frac{1}{A}\)+ \(\frac{\sqrt{a}+1}{8}=0\)

Answer 1

[Trần Đức Thắng]

Đặt B  = \(\left(\frac{\left(a+3\sqrt{a}+1\right)\left(\sqrt{a}+1\right)-\left(a+\sqrt{a}\right)\left(\sqrt{a}+2\right)}{\left(\text{\sqrt{a}+2}\right)\left(a-1\right)}\right)\)  ($\sqrt{ a}$ + 2 là căn a )

\(=\frac{a\sqrt{a}+a+3a+3\sqrt{a}+\sqrt{a}+1-a\sqrt{a}-2a-a-2\sqrt{a}}{\left(\sqrt{a}+2\right)\left(a-1\right)}\)

\(\frac{a+3\sqrt{a}+2}{\left(\sqrt{a}+2\right)\left(a-1\right)}=\frac{\left(\sqrt{a}+2\right)\left(\sqrt{a}+1\right)}{\left(\sqrt{a}+2\right)\left(a-1\right)}=\frac{\sqrt{a}+1}{a-1}\)(vì a - 1 = (căn a - 1 ) (căn a + 1 ) )

Dặt \(C=\frac{1}{\sqrt{a}+1}-\frac{1}{\sqrt{a}-1}=\frac{\sqrt{a}-1-\sqrt{a}-1}{a-1}=-\frac{2}{a-1}\)

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

Answer 2

[Trần Đức Thắng]

Rut gon lai là (-1-căn a) / 2 

b,a = 3