a: Ta có: \(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{x-4}\)

\(=\dfrac{x+4\sqrt{x}+4-3\sqrt{x}+6+12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}\)

\(=\dfrac{x+\sqrt{x}+22}{x-4}\)

d: Ta có: \(D=\dfrac{1}{\sqrt{x}+3}-\dfrac{\sqrt{x}}{3-\sqrt{x}}+\dfrac{2\sqrt{x}-12}{x-9}\)

\(=\dfrac{\sqrt{x}-3+x+3\sqrt{x}+2\sqrt{x}-12}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(=\dfrac{x+6\sqrt{x}-15}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(B=\dfrac{x+3+2\left(\sqrt{x}-3\right)+\sqrt{x}+3}{x-9}\)

\(=\dfrac{x+\sqrt{x}+6+2\sqrt{x}-6}{x-9}=\dfrac{x+3\sqrt{x}}{x-9}\)

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

\(B=\dfrac{x+3}{x-9}+\dfrac{2}{3+\sqrt{x}}-\dfrac{1}{3-\sqrt{x}}\\ B=\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2}{\sqrt{x}+3}+\dfrac{1}{\sqrt{x}-3}\\ B=\dfrac{x+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{2\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{x+3+2\sqrt{x}-6+\sqrt{x}+3}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{x+3\sqrt{x}}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\\ B=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}\left(\text{đ}pcm\right)\)

\(B=\dfrac{\sqrt{x}+3}{\sqrt{x}-3}+\dfrac{\sqrt{x}-3}{\sqrt{x}+3}-\dfrac{36}{x-9}\)

\(=\dfrac{\left(\sqrt{x}+3\right)^2+\left(\sqrt{x}-3\right)^2-36}{x-9}\)

\(=\dfrac{x+6\sqrt{x}+9+x-6\sqrt{x}+9-36}{x-9}\)

\(=\dfrac{2x-18}{x-9}=\dfrac{2\left(x-9\right)}{x-9}=2\)

ĐKXĐ : \(x\ge0;x\ne9\)

Ta có : \(B=\dfrac{\left(\sqrt{x}+3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}+\dfrac{\left(\sqrt{x}-3\right)^2}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}-\dfrac{36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(B=\dfrac{x+6\sqrt{x}+9+x-6\sqrt{x}+9-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2x-18}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{2\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=2\)

1) Thay x=16 vào biểu thức ta có:

 \(A=\dfrac{\sqrt{x}}{\sqrt{x+3}}=\dfrac{\sqrt{16}}{\sqrt{16}+3}=\dfrac{4}{4+3}=\dfrac{4}{7}\)

2) \(A+B=\dfrac{\sqrt{x}}{\sqrt{x}+3}+\dfrac{2\sqrt{x}}{\sqrt{x}-3}-\dfrac{3x+9}{x-9}\\ \Rightarrow A+B=\dfrac{\sqrt{x}\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{2\sqrt{x}\left(\sqrt{x}+3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}-\dfrac{3x+9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(\Rightarrow A+B=\dfrac{x-3\sqrt{x}+2x+6\sqrt{x}-3x-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(\Rightarrow A+B=\dfrac{3\sqrt{x}-9}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(\Rightarrow A+B=\dfrac{3\left(\sqrt{x}-3\right)}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\)

\(\Rightarrow A+B=\dfrac{3}{\sqrt{x}+3}\)

1: Thay x=16 vào A, ta được:

\(A=\dfrac{4}{4+3}=\dfrac{4}{7}\)

a: \(A=\dfrac{2x-6\sqrt{x}+x+4\sqrt{x}+3-3+11\sqrt{x}}{x-9}\)

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

a) \(B=\dfrac{\sqrt{x}}{\sqrt{x}-3}+\dfrac{8\sqrt{x}+24}{x-9}\)

\(=\dfrac{\sqrt{x}\left(\sqrt{x}+3\right)+8\sqrt{x}+24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

\(=\dfrac{x+3\sqrt{x}+8\sqrt{x}+24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}\)

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

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

b) Mình không biết làm bạn thông cảm.

cảm ơn bn ạ