Bài 1: 

a: \(A=\dfrac{x^2-3+x+3}{\left(x-3\right)\left(x+3\right)}\cdot\dfrac{x+3}{x}=\dfrac{x\left(x+1\right)}{x\left(x-3\right)}=\dfrac{x+1}{x-3}\)

b: Để A=3 thì 3x-9=x+1

=>2x=10

hay x=5

Bài 2: 

a: \(A=\dfrac{x+x-2-2x-4}{\left(x-2\right)\left(x+2\right)}:\dfrac{x+2-x}{x+2}\)

\(=\dfrac{-6}{x-2}\cdot\dfrac{1}{2}=\dfrac{-3}{x-2}\)

b: Để A nguyên thì \(x-2\in\left\{1;-1;3;-3\right\}\)

hay \(x\in\left\{3;1;5;-1\right\}\)

Bạn viết lại đề bài nhé!

Giúp tớ với

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

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

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

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

\(P=\dfrac{-x^4+2x^3-2x+1}{4x^2-1}+\dfrac{8x^2-4x+2}{8x^3+1}\)

\(=\dfrac{\left(1-x^2\right)\left(1+x^2\right)+2x\left(x^2-1\right)}{4x^2-1}+\dfrac{2\left(4x^2-2x+1\right)}{\left(2x+1\right)\left(4x^2-2x+1\right)}\)

\(=\dfrac{\left(1-x^2\right)\left(1+x^2-2x\right)}{4x^2-1}+\dfrac{2}{2x+1}\)

\(=\dfrac{\left(1-x^2\right)\left(x^2-2x+1\right)+4x-2}{4x^2-1}\)

Với \(x\ne1\)ta có 

\(P=\left(\frac{4}{x-1}-\frac{7x+5}{x^3-1}\right):\left(1-\frac{x-4}{x^2+x+1}\right)\)

\(=\left[\frac{4x^2+4x+4-7x-5}{\left(x-1\right)\left(x^2+x+1\right)}\right]:\left(\frac{x^2+x+1-x-4}{x^2+x+1}\right)\)

\(=\frac{4x^2-3x-1}{\left(x-1\right)\left(x^2+x+1\right)}:\frac{x^2-3}{x^2+x+1}=\frac{4x+1}{x^2-3}\)

a: \(A=\left(\dfrac{4}{x}-1\right):\left(1-\dfrac{x-3}{x^2+x+1}\right)\)

\(=\dfrac{4-x}{x}:\dfrac{x^2+x+1-x+3}{x^2+x+1}\)

\(=\dfrac{4-x}{x}\cdot\dfrac{x^2+x+1}{x^2+4}=\dfrac{\left(4-x\right)\left(x^2+x+1\right)}{x\left(x^2+4\right)}\)

b: x^4-7x^2-4x+20=0

=>(x-2)^2(x^2+4x+5)=0

=>x=2

Khi x=2 thì \(A=\dfrac{\left(4-2\right)\left(4+2+1\right)}{2\left(4+4\right)}=\dfrac{7}{8}\)