Question

1. Cho biểu thức B :

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

a. Rút gọn B

b. Tính giá trị K khi a= \(3+2\sqrt{2}\)

c. Tìm các giá trị của a sao cho K<0

Answer 1

Lời giải:

ĐK: \(a>0; a\neq 1\)

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

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

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

\(=\frac{a-1}{a-\sqrt{a}}: \frac{\sqrt{a}+1}{a-1}=\frac{a-1}{a-\sqrt{a}}.\frac{a-1}{\sqrt{a}+1}=\frac{(a-1)^2}{\sqrt{a}(\sqrt{a}-1)(\sqrt{a}+1)}=\frac{(a-1)^2}{\sqrt{a}(a-1)}=\frac{a-1}{\sqrt{a}}\)

b) Ta có:
\(a=3+2\sqrt{2}=2+1+2\sqrt{2}=(\sqrt{2}+1)^2\)

\(\Rightarrow K=\frac{3+2\sqrt{2}-1}{\sqrt{2}+1}=\frac{2+2\sqrt{2}}{\sqrt{2}+1}=\frac{2(1+\sqrt{2})}{\sqrt{2}+1}=2\)

c) \(K< 0\leftrightarrow \frac{a-1}{\sqrt{a}}< 0\Leftrightarrow a-1< 0\) (do \(\sqrt{a}>0\))

\(\Leftrightarrow a< 1\)

Vậy \(0< a< 1\)