ĐK: x khác 1;-1
a) Thay x = 25 (TMDK) vào A, ta có:
\(A=\dfrac{2\sqrt{25}-4}{\sqrt{25}-1}=\dfrac{3}{2}\)
b) \(B=\dfrac{\sqrt{x}}{\sqrt{x}-1}+\dfrac{3}{\sqrt{x}+1}-\dfrac{6\sqrt{x}-4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)+3\left(\sqrt{x}-1\right)-6\sqrt{x}+4}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\)
= \(\dfrac{x-2\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}=\dfrac{\sqrt{x}-1}{\sqrt{x}+1}\)
c) Xét A.B = \(\dfrac{2\sqrt{x}-4}{\sqrt{x}-1}.\dfrac{\sqrt{x}-1}{\sqrt{x}+1}=\dfrac{2\sqrt{x}-4}{\sqrt{x}+1}\) = \(2-\dfrac{6}{\sqrt{x+1}}< 2< 5\)
Đúng 1
Bình luận (0)
