[2x-2=0=>x=1

x-1=0=>x=1

x+1=0=>x=-1

5=0=>x=5

ĐKXĐ: \(x\ge0;x\ne9\)

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

\(=\left[\dfrac{\sqrt{x}-3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}+\dfrac{3}{\left(\sqrt{x}+3\right)\left(\sqrt{x}-3\right)}\right]:\dfrac{1}{\sqrt{x}-3}\)

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

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

ĐKXĐ: \(\left\{{}\begin{matrix}x>=0\\x< >9\end{matrix}\right.\)

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

a: ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

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

a: ĐKXĐ: 

b: 

a) A đc xác định <=>2x+4\(\left\{{}\begin{matrix}2x+4\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne2\end{matrix}\right.\)

câu b bn quy đòng mẫu là đc

a) ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

b) Ta có: \(A=\dfrac{x}{2x+4}+\dfrac{3x+2}{x^2-4}\)

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

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

\(=\dfrac{x^2-2x+6x+4}{2\left(x+2\right)\left(x-2\right)}\)

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

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

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

c) Để A=0 thì \(\dfrac{x+2}{2\left(x-2\right)}=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2(Không thỏa mãn ĐKXĐ)

Vậy: Không có giá trị nào của x để A=0

đk : x >= 0 ; x khác 1 

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

B xác định \(< =>\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

ĐKXĐ:\(\left\{{}\begin{matrix}x\ge0\\\sqrt{x}-1\ne0\\\sqrt{x}+1\ne0\left(luôn.đúng\right)\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\\sqrt{x}\ne1\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ge0\\x\ne1\end{matrix}\right.\)

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

a) ĐKXĐ: \(x\notin\left\{1;-1\right\}\)

b) Ta có: \(B=\left(\dfrac{x-2}{2x-2}+\dfrac{3}{2x-2}-\dfrac{x+3}{2x+2}\right):\left(1-\dfrac{x-3}{x+1}\right)\)

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

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

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

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

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

\(=\dfrac{1}{2}\)

Vậy: Khi x=2005 thì \(B=\dfrac{1}{2}\)

a/

Để biểu thức được xác định

\(=>\left\{{}\begin{matrix}2x-2\ne0\\2x+2\ne0\\x+1\ne0\end{matrix}\right.\)

\(\odot2x-2\ne0\)

\(2x\ne2\)

\(x\ne1\)

\(\odot2x+2\ne0\)

\(2x\ne-2\)

\(x\ne-1\)

\(\odot x+1\ne0\)

\(x\ne-1\)

Vậy điều kiện xác định của bt là: \(x\ne-1;x\ne\pm2\)

Đề bài là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2-\left(x+2\right)^2}\) hay là \(B=\dfrac{\left(x-1\right)^2-4}{\left(2x+1\right)^2}-\left(x+2\right)^2?\)

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

viết lại biểu thức 

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

\(\Rightarrow\) ĐKXĐ: \(x\ne\pm1\)

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

c) Thay \(x=-3;x=1\) vào (2) ta có: \(\left\{{}\begin{matrix}B=\dfrac{-3-3}{3.\left(-3\right)-3}=\dfrac{1}{2}\\B=\dfrac{1-3}{3.1-3}=0\end{matrix}\right.\)

d) \(B=5\Rightarrow\dfrac{x-3}{3x-3}=5\Leftrightarrow x-3=15x-15\Leftrightarrow x=\dfrac{6}{7}\)

\(a,A=\dfrac{x+1+2-2x+5-x}{\left(1-x\right)\left(x+1\right)}\cdot\dfrac{\left(1-x\right)\left(x+1\right)}{2x-1}\left(x\ne1;x\ne-1;x\ne\dfrac{1}{2}\right)\\ A=\dfrac{8-2x}{2x-1}\\ b,A>0\Leftrightarrow\dfrac{8-2x}{2x-1}>0\\ \Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}8-2x>0\\2x-1>0\end{matrix}\right.\\\left\{{}\begin{matrix}8-2x< 0\\2x-1< 0\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\left\{{}\begin{matrix}x< 4\\x>\dfrac{1}{2}\end{matrix}\right.\\\left\{{}\begin{matrix}x>4\\x< \dfrac{1}{2}\end{matrix}\right.\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}\dfrac{1}{2}< x< 4\\x\in\varnothing\end{matrix}\right.\Leftrightarrow\dfrac{1}{2}< x< 4\)