\(a,C=\dfrac{2x^2-x-x-1+2-x^2}{x-1}\left(x\ne1\right)\\ C=\dfrac{x^2-2x+1}{x-1}=\dfrac{\left(x-1\right)^2}{x-1}=x-1\\ b,D=\dfrac{1+\sqrt{a}}{\sqrt{a}\left(\sqrt{a}-1\right)}\cdot\dfrac{\left(\sqrt{a}-1\right)^2}{\sqrt{a}+1}\left(a>0;a\ne1\right)\\ D=\dfrac{\sqrt{a}-1}{\sqrt{a}}\)

Có 

\(\left(\dfrac{\dfrac{x}{x+1}}{\dfrac{x^2}{x^2+x+1}}-\dfrac{2x+1}{x^2+x}\right)\dfrac{x^2-1}{x-1}\)ĐK : \(x\ne\pm1\)

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

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

à xin lỗi mình nhầm dòng cuối 

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

Để biểu thức trên nhận giá trị dương khi 

\(\dfrac{\left(x+1\right)\left(x-2\right)}{x}>0\)bạn tự xét TH cả tử và mẫu nhé, mình đánh trên này bị lỗi 

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

= 2x+1 - x² + 2x²-3x-1

= -x + x²

ĐKXĐ : x ≠ 0 ; x ≠ \(-\dfrac{1}{2}\)

\(\Rightarrow2x+1-x^2=2x^2-3x-1\)

\(\Leftrightarrow-3x^2+5x+2=0\)

\(\Leftrightarrow-3x^2-x+6x+2=0\)

\(\Leftrightarrow-x\left(3x+1\right)+2\left(3x+1\right)=0\)

\(\Leftrightarrow\left(3x+1\right)\left(-x+2\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=-\dfrac{1}{3}\\x=2\end{matrix}\right.\) (N)

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

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

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

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

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

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

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

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

ĐKXĐ: \(x\notin\left\{1;\dfrac{1}{2}\right\}\)

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

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

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

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

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

a) Ta có: \(\dfrac{2x^2-2x}{x-1}\)

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

=2x

b) Ta có: \(\dfrac{x^2+2x+1}{3x^2+3x}\)

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

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

c) Ta có: \(\dfrac{x}{3x-3}+\dfrac{1}{x^2-1}\)

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

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

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

a, \(\dfrac{2x^2-2x}{x-1}=\dfrac{2x\left(x-1\right)}{x-1}=2x\) ( đk : \(x\ne1\) )

b,\(\dfrac{x^2+2x+1}{3x^2+3x}=\dfrac{\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{x+1}{3x}\) ( đk : \(x\ne-1\) )

c

=

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

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

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

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

ĐKXĐ: ...

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

\(=\dfrac{10}{x+1}-\dfrac{x-1}{x+1}\)

\(=\dfrac{11-x}{x+1}\)

ĐK: \(x\ne-1\)

Rút gọn:

\(\dfrac{10}{x+1}-\dfrac{x^2-1}{x^2+2x+1}\\ =\dfrac{10\left(x+1\right)-\left(x^2-1\right)}{\left(x+1\right)^2}\\ =\dfrac{\left(x+1\right).\left(10-x+1\right)}{\left(x+1\right)^2}=\dfrac{\left(x+1\right).\left(11-x\right)}{\left(x+1\right)^2}\\ =\dfrac{11-x}{x+1}\)

` @ \color{Red}{m}`

` \color{lightblue}{Answer}`  

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

__

\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\\ =\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2x-6}{2x\left(x+3\right)}\\ =\dfrac{3x-\left(2x-6\right)}{2x\left(x+3\right)}\\ =\dfrac{3x-2x+6}{2x\left(x+3\right)}\\ =\dfrac{x+6}{2x\left(x+3\right)}\)

__

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

\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\left(dkxd:x\ne\pm1\right)\)

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

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

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

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

========================

\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\left(dkxd:x\ne\pm3;x\ne0\right)\)

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

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

\(=\dfrac{3x-2x+6}{2x\left(x+3\right)}\)

\(=\dfrac{x+6}{2x^2+6x}\)

==========================

\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\left(dkxd:x\ne\pm1\right)\)

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

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

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

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

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