Ta có: \(B=\left(\dfrac{1}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\right)\cdot\dfrac{4\sqrt{x}+12}{\sqrt{x}}\)

\(=\dfrac{\sqrt{x}-3+\sqrt{x}+3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\cdot\dfrac{4\left(\sqrt{x}+3\right)}{\sqrt{x}}\)

\(=\dfrac{8\sqrt{x}}{\sqrt{x}\left(\sqrt{x}-3\right)}\)

\(=\dfrac{8}{\sqrt{x}-3}\)

gõ latex đi b=)

\(A=\sqrt{x}+1\) (đã thu gọn)

\(B=\dfrac{4\sqrt{x}}{x+4}\) (đã thu gọn)

\(A=x-\sqrt{x}+1=\sqrt{x}\cdot\sqrt{x}-\sqrt{x}+1=\sqrt{x}\left(\sqrt{x}-1\right)+1\)

\(A=\dfrac{3}{2\sqrt{x}}\) (đã thu gọn)

\(A=\dfrac{3}{\sqrt{x}+3}\) (đã thu gọn)

\(A=1-\sqrt{x}\) (đã thu gọn)

\(A=x-2\sqrt{x}-1=\sqrt{x}\left(\sqrt{x}-2\right)-1\)

\(A=\left(\frac{4\sqrt{x}}{\sqrt{x}+2}+\frac{8x}{4-x}\right):\left(\frac{\sqrt{x}-1}{x+2\sqrt{x}}-\frac{2}{\sqrt{x}}\right)\)

\(=\left(\frac{4\sqrt{x}\left(\sqrt{x}-2\right)-8x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\right):\left(\frac{\sqrt{x}-1}{\sqrt{x}\left(\sqrt{x}+2\right)}-\frac{2}{\sqrt{x}}\right)\)

\(=\frac{4x-8\sqrt{x}-8x}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}:\frac{\sqrt{x}-1-2\left(\sqrt{x}+2\right)}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

\(=\frac{-4x-8\sqrt{x}}{\left(\sqrt{x}-2\right)\cdot\left(\sqrt{x}+2\right)}:\frac{\sqrt{x}-1-2\sqrt{x}-4}{\sqrt{x}\left(\sqrt{x}+2\right)}\)

\(=\frac{-4\sqrt{x}\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\cdot\frac{\sqrt{x}\left(\sqrt{x}+2\right)}{-\sqrt{x}-5}=\frac{4x\left(\sqrt{x}+2\right)}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+5\right)}\)

Mình cần gấp nha mn 😭😭 

1) Ta có: \(\frac{x+6\sqrt{x}+9}{x-9}=\frac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\frac{\sqrt{x}+3}{\sqrt{x}-3}\)