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

\(=\dfrac{2x-2\sqrt{x}-\sqrt{x}+1}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2\sqrt{x}\left(\sqrt{x}-1\right)-\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{\left(\sqrt{x}-1\right)\left(2\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(\sqrt{x}+1\right)}\\ =\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\left(\text{đ}pcm\right)\)

A+B

\(=\dfrac{\sqrt{x}+1}{\sqrt{x}-1}+\dfrac{\sqrt{x}-1}{\sqrt{x}+1}-\dfrac{3\sqrt{x}+1}{x-1}\)

\(=\dfrac{x+2\sqrt{x}+1+x-2\sqrt{x}+1-3\sqrt{x}-1}{x-1}\)

\(=\dfrac{2x-3\sqrt{x}+1}{x-1}=\dfrac{2\sqrt{x}-1}{\sqrt{x}+1}\)

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) ĐKXĐ: \(x\ge0,x\ne9\)

\(B=\dfrac{x+3\sqrt{x}+2\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{x+5\sqrt{x}-24}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\left(\sqrt{x}-3\right)\left(\sqrt{x}+8\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{\sqrt{x}+8}{\sqrt{x}+3}\)

b) \(\dfrac{\sqrt{x-1}}{\sqrt{x}+2}=0\left(đk:x\ge0\right)\)\(\Leftrightarrow\sqrt{x-1}=0\Leftrightarrow x-1=0\Leftrightarrow x=1\left(tm\right)\)

1, Thay x = 16 vào ta được \(A=\dfrac{4}{4+3}=\dfrac{4}{7}\)

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

Ta có đpcm 

A

Thay x=16 vào biểu thức A , ta có :

A= 

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:

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

\(a,\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(dkxd:a\ne9,a\ge0\right)\)

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

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

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

\(=\dfrac{11}{a-9}\)

\(b,\dfrac{x+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x+\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(dkxd:x\ge0,x\ne1\right)\)

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

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

\(=\dfrac{x+2+x-1-x-\sqrt{x}-1}{x\sqrt{x}-1}\)

\(=\dfrac{x-\sqrt{x}}{x\sqrt{x}-1}\)

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

\(\dfrac{\sqrt{a}}{\sqrt{a}-3}-\dfrac{3}{\sqrt{a}+3}-\dfrac{a-2}{a-9}\left(\text{đ}k\text{x}\text{đ}:a\ge0;a\ne9\right)\\ =\dfrac{\sqrt{a}\left(\sqrt{a}+3\right)}{\left(\sqrt{a-3}\right)\left(\sqrt{a+3}\right)}-\dfrac{3\left(\sqrt{a}-3\right)}{\left(\sqrt{a}+3\right)\left(\sqrt{a}-3\right)}-\dfrac{a-2}{\left(\sqrt{a}+3\right)\left(\sqrt{a-3}\right)}\\ =\dfrac{a+3\sqrt{a}-\left(3\sqrt{a}-9\right)-\left(a-2\right)}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\\ =\dfrac{a+3\sqrt{a}-3\sqrt{a}+9-a+2}{\left(\sqrt{a}-3\right)\left(\sqrt{a}+3\right)}\\ =\dfrac{11}{\left(\sqrt{a}-3\right)\left(\sqrt{a+3}\right)}\)

\(b,\dfrac{a+2}{x\sqrt{x}-1}+\dfrac{\sqrt{x}+1}{x\sqrt{x}+1}-\dfrac{1}{\sqrt{x}-1}\left(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne1\right)\\ =\dfrac{x+2}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}+\dfrac{\left(\sqrt{x}+1\right)\left(\sqrt{x}-1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}-\dfrac{1\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\\ =\dfrac{x+2+\left(\sqrt{x}+1\right)\left(\sqrt{x-1}\right)-\left(x+\sqrt{x}+1\right)}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x+1}\right)}\\ =\dfrac{x+2+x-1-x-\sqrt{x}-1}{\left(\sqrt{x}-1\right)\left(x+\sqrt{x}+1\right)}\)

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

\(B=\dfrac{x+6\sqrt{x}+9-x+6\sqrt{x}-9-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{12\sqrt{x}-36}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{12\left(\sqrt{x}-3\right)}{\left(\sqrt{x}-3\right)\left(\sqrt{x}+3\right)}=\dfrac{12}{\sqrt{x}+3}\)

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

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

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

A=\(A=\dfrac{\sqrt{x}+2}{\sqrt{x}-2}-\dfrac{3}{\sqrt{x}+2}+\dfrac{12}{\left(\sqrt{x}-2\right)\left(\sqrt{x}+2\right)}A=\dfrac{\sqrt{x}+2.\left(\sqrt{x}+2\right)-3.\left(\sqrt{x}-2\right)+12}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}A=\dfrac{\sqrt{x}+2\sqrt{x}+4-3\sqrt{x}+6+12}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}A=\dfrac{22}{\left(\sqrt{x}-2\right).\left(\sqrt{x}+2\right)}\)