a, Thay x = 16 ta được
\(A=\dfrac{16}{4-3}=16\)
b, Với x > 0 ; x khác 9
\(B=\dfrac{2x-3-\left(\sqrt{x}-3\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}=\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}\)
Vậy ta có đpcm
c, \(A-B< 0\Leftrightarrow\dfrac{x}{\sqrt{x}-3}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-3}< 0\Rightarrow\sqrt{x}< 3\Leftrightarrow x< 9\)
Kết hợp đk vậy 0 < x < 9
1) Thay `x=16` vào A ta có:
\(A=\dfrac{16}{\sqrt{16}-3}=\dfrac{16}{4-3}=16\)
2)
\(B=\dfrac{2x-3}{x-3\sqrt{x}}-\dfrac{1}{\sqrt{x}}\\ =\dfrac{2x-3}{\sqrt{x}\left(\sqrt{x}-3\right)}-\dfrac{\sqrt{x}-3}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{2x-3-\sqrt{x}+3}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{2x-\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{\sqrt{x}\left(2\sqrt{x}-1\right)}{\sqrt{x}\left(\sqrt{x}-3\right)}\\ =\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}\)
3)
\(A-B=\dfrac{x}{\sqrt{x}-3}-\dfrac{2\sqrt{x}-1}{\sqrt{x}-3}=\dfrac{x-2\sqrt{x}+1}{\sqrt{x}-3}=\dfrac{\left(\sqrt{x}-1\right)^2}{\sqrt{x}-3}\)
Vì: \(\left(\sqrt{x}-1\right)^2\ge0\) nên để \(A-B< 0\Rightarrow\sqrt{x}-3< 0\Leftrightarrow x< 9\)
Kết hợp với đk: 0 < x < 9
