A=\(\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
A=\(\dfrac{\left(\sqrt{x}+1\right)^2+\left(\sqrt{x}-1\right)^2-3\sqrt{x}-1}{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}\)
\(A=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{2x-3\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
\(A=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)
b)
Ta có: \(x=4+2\sqrt{3}\)
\(x=\left(1+\sqrt{3}\right)^2\)
Thay \(x=\left(1+\sqrt{3}\right)^2\) (TMĐK) vào A ta có:
\(A=\)\(\dfrac{2\sqrt{\left(1+\sqrt{3}\right)^2}-1}{\sqrt{\left(1+\sqrt{3}\right)^2}+1}\)
A=\(\dfrac{2\left(1+\sqrt{3}\right)-1}{1+\sqrt{3}+1}\)
A=\(\dfrac{2+2\sqrt{3}-1}{2+\sqrt{3}}\)
A=\(\dfrac{1+2\sqrt{3}}{2+\sqrt{3}}\)
A= \(-4+3\sqrt{3}\)
c) \(A=\dfrac{1}{2}\)
⇔\(\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{1}{2}\)
⇔\(4\sqrt{x}-2=\sqrt{x}+1\)
⇔\(3\sqrt{x}=3\)
⇔\(\sqrt{x}=1\)
⇔\(x=1\)
Vậy với x=1 thì A=\(\dfrac{1}{2}\)
