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

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

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

b: Để A là số nguyên thì \(\sqrt{x}-1⋮\sqrt{x}+1\)

=>\(\sqrt{x}+1-2⋮\sqrt{x}+1\)

=>căn x+1 thuộc {1;2}

=>căn x thuộc {0;1}

mà x<>1

nên x=0

`A=(1/(x-sqrtx)+1/(sqrtx-1)):(sqrtx+1)/(sqrtx-1)^2`

`=((sqrtx+1)/(x-sqrtx)).(sqrtx-1)^2/(sqrtx+1)`

`=(sqrtx-1)^2/(x-sqrtx)`

`=(sqrtx-1)/sqrtx`

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

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

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

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

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

Ta có: \(P=\dfrac{x^2+\sqrt{x}}{x-\sqrt{x}+1}-\dfrac{2\left(x+\sqrt{x}\right)}{\sqrt{x}}+\dfrac{2\left(x-1\right)}{\sqrt{x}-1}\)

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

\(=x+\sqrt{x}\)

Mẫu chung là gì cơ ạ?