\(a,DKXD:x\ge0\)

\(b,A=\sqrt{x-\sqrt{x^2-4x+4}}\)

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

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

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

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

\(=\sqrt{2}\)

a: ĐKXĐ: x<>0; x<>3

b: \(P=\dfrac{3\left(x-3\right)}{x\left(x-3\right)}=\dfrac{3}{x}\)

ĐKXĐ: x<>0; x<>2; x<>1; x<>-1

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

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

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

a: ĐKXĐ: \(x\ne-1\)

b: \(B=\dfrac{5}{x+1}+\dfrac{10}{x^2-x+1}-\dfrac{15}{x^3-1}\)

\(=\dfrac{5x^2-5x+5+10x+10-15}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{5x^2+5x}{\left(x+1\right)\left(x^2-x+1\right)}\)

\(=\dfrac{5x}{x^2-x+1}\)

ĐKXĐ: x>=0; x<>4

\(B=\dfrac{x}{x-4}-\dfrac{1}{2-\sqrt{x}}+\dfrac{1}{\sqrt{x}+2}\)

\(=\dfrac{x+\sqrt{x}+2+\sqrt{x}-2}{x-4}=\dfrac{x+2\sqrt{x}}{x-4}\)

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

`a)->` ĐKXĐ : `x>=0;x\ne1`

`b)` Ta có :

`P=(\sqrtx)/(\sqrtx-1)-(2\sqrtx)/(\sqrtx+1)+(x-3)/(x-1)`

`P=(\sqrtx(\sqrtx+1)-2\sqrtx(\sqrtx-1)+x-3)/(x-1)`

`P=(x+\sqrtx-2x+2\sqrtx+x-3)/(x-1)`

`P=(3\sqrtx-3)/(x-1)`

`P=(3(\sqrtx-1))/((\sqrtx-1)(\sqrtx+1))`

`P=3/(\sqrtx+1)`

Vậy `P=3/(\sqrtx+1)` khi `x>=0;x\ne1`

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

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

Bổ sung \(\text{đ}k\text{x}\text{đ}:x\ge0;x\ne1\)

ĐKXĐ: a>=0; a<>9

\(Q=\dfrac{3}{\sqrt{a}-3}+\dfrac{2}{\sqrt{a}+3}-\dfrac{a-5\sqrt{a}-3}{9-a}\)

\(=\dfrac{3\left(\sqrt{a}+3\right)+2\left(\sqrt{a}-3\right)+a-5\sqrt{a}-3}{a-9}\)

\(=\dfrac{3\sqrt{a}+9+2\sqrt{a}-6+a-5\sqrt{a}-3}{a-9}=\dfrac{a}{a-9}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge0\\x\ne2\end{matrix}\right.\)

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

\(M=\dfrac{-8\sqrt{x}}{x-4}\)

\(M< 0\Leftrightarrow-\dfrac{8\sqrt{x}}{x-4}< 0\Leftrightarrow x-4>0\Leftrightarrow x>4\)