a/ ĐKXĐ : \(\left\{{}\begin{matrix}x-1\ne0\\x+1\ne0\end{matrix}\right.\) \(\Leftrightarrow\left\{{}\begin{matrix}x\ne1\\x\ne-1\end{matrix}\right.\)

Vậy..

b/ Ta có :

\(C=\left(\frac{2x+1}{x-1}+\frac{8}{x^2-1}-\frac{x-1}{x+1}\right).\frac{x^2-1}{5}\)

\(=\left(\frac{2x+1}{x-1}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{x-1}{x+1}\right).\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\left(\frac{\left(2x+1\right)\left(x+1\right)}{\left(x-1\right)\left(x+1\right)}+\frac{8}{\left(x-1\right)\left(x+1\right)}-\frac{\left(x-1\right)^2}{\left(x+1\right)\left(x-1\right)}\right).\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{2x^2+2x+x+1+8-x^2+2x-1}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{x^2+5x+8}{\left(x-1\right)\left(x+1\right)}.\frac{\left(x-1\right)\left(x+1\right)}{5}\)

\(=\frac{\left(x+\frac{5}{2}\right)^2+\frac{7}{4}}{5}\)

Vậy...

c/ Với mọi x ta có :

\(\left\{{}\begin{matrix}\left(x+\frac{5}{2}\right)^2+\frac{7}{4}>0\\5>0\end{matrix}\right.\)

\(\Leftrightarrow\frac{\left(x+\frac{5}{2}\right)^2+\frac{7}{4}}{5}>0\)

\(\Leftrightarrow C>0\left(đpcm\right)\)

$\textbf{a)}$

\[\left(x+3-\frac1{x+3}\right)\cdot\frac{x+3}{x+4}\]

$\text{ĐKXĐ: }x\ne-3,\,-4.$

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

$=\dfrac{(x+2)(x+4)}{x+4}$

$=x+2.$

$\textbf{b)}$

$\left(2x-4-\frac{x-12}{3x+4}\right)\cdot\left(3x-2-\frac{10}{2x+1}\right)$

$\text{ĐKXĐ: }x\ne-\dfrac43,\,-\dfrac12.$

$=\dfrac{(2x-4)(3x+4)-(x-12)}{3x+4}\cdot\dfrac{(3x-2)(2x+1)-10}{2x+1}$

$=\dfrac{6x^2-5x-4}{3x+4}\cdot\dfrac{6x^2-x-12}{2x+1}.$

a)

\(\frac{7x-3}{x-1}=\frac{2}{3}\\ \Leftrightarrow\frac{21x-9}{3\cdot\left(x-1\right)}-\frac{2x-2}{3\cdot\left(x-1\right)}=0\\ \Leftrightarrow21x-9-2x+2=0\\ \Leftrightarrow19x-7=0\\ \Rightarrow x=\frac{7}{19}\)

Vậy \(x=\left\{\frac{7}{19}\right\}\) là nghiệm của phương trình.

b)

\(\frac{1}{x-2}+3=\frac{3-x}{x-2}\\ \Leftrightarrow\frac{1}{x-2}+\frac{3x-6}{x-2}-\frac{3-x}{x-2}=0\\ \Leftrightarrow1+3x-6-3+x=0\\ \Leftrightarrow4x-8=0\\ \Rightarrow x=\frac{8}{4}=2\)

Mà \(ĐKXĐ:x\ne2\\ \Rightarrow x\in\varnothing\)

Hay phương trình vô nghiệm.

c)

\(\frac{8-x}{x-7}-8=\frac{1}{x-7}\\ \Leftrightarrow\frac{8-x}{x-7}-\frac{8x-56}{x-7}-\frac{1}{x-7}=0\\ \Leftrightarrow8-x-8x+56-1=0\\ \Leftrightarrow63-9x=0\\ \Rightarrow x=\frac{63}{9}=7\)

Mà \(ĐKXĐ:x\ne7\\ \Rightarrow x\in\varnothing\)

Hay phương trình vô nghiệm.

d)

\(\frac{x+5}{x-5}-\frac{x-5}{x+5}=\frac{20}{x^2-25}\\ \Leftrightarrow\frac{\left(x+5\right)^2}{x^2-25}-\frac{\left(x-5\right)^2}{x^2-25}-\frac{20}{x^2-25}=0\\ \Leftrightarrow\left(x+5\right)^2-\left(x-5\right)^2-20=0\\ \Leftrightarrow x^2+10x+25-x^2+10x-25-20=0\\ \Leftrightarrow20x-20=0\\ \Rightarrow x=1\)

Vậy \(x=1\) là nghiệm của phương trình.

e)

\(\frac{x}{2\cdot\left(x-3\right)}+\frac{x}{2\cdot\left(x+1\right)}=\frac{2x}{\left(x+1\right)\cdot\left(x-3\right)}\\ \Leftrightarrow\frac{x^2+x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}+\frac{x^2-3x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}-\frac{4x}{2\cdot\left(x+1\right)\cdot\left(x-3\right)}=0\\ \Leftrightarrow x^2+x+x^2-3x-4x=0\\ \Leftrightarrow2x^2-6x=0\\ \Leftrightarrow2x\cdot\left(x-3\right)=0\\ \Rightarrow\left[{}\begin{matrix}2x=0\\x-3=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=0\\x=3\end{matrix}\right.\)

Mà \(ĐKXĐ:x\ne\left\{-1;3\right\}\\ \Rightarrow x=0\)

Vật \(x=0\) là nghiệm của phương trình.

f)

\(\frac{1}{x}+\frac{1}{x+10}=\frac{1}{12}\\ \Leftrightarrow\frac{12x+120}{12x\cdot\left(x+10\right)}+\frac{12x}{12x\cdot\left(x+10\right)}-\frac{x^2+10x}{12x\cdot\left(x+10\right)}=0\\ \Leftrightarrow12x+120+12x-x^2-10x=0\\ \Leftrightarrow14x+120-x^2=0\\ \Leftrightarrow x^2-14x-120=0\\ \Leftrightarrow x^2+6x-20x-120=0\\ \Leftrightarrow x\cdot\left(x+6\right)-20\cdot\left(x+6\right)=0\\ \Leftrightarrow\left(x-20\right)\cdot\left(x+6\right)=0\\ \Rightarrow\left[{}\begin{matrix}x-20=0\\x+6=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=20\\x=-6\end{matrix}\right.\)

Vậy \(x=\left\{20;-6\right\}\) là ngiệm của phương trình.