\(ĐKXĐ:\hept{\begin{cases}x\ne2\\x\ne4\end{cases}}\)

\(\frac{x+3}{x-4}+\frac{x-1}{x-2}=\frac{2}{6x-8-x^2}\)

\(\Leftrightarrow\frac{x+3}{x-4}+\frac{x-1}{x-2}+\frac{2}{\left(x-2\right)\left(x-4\right)}=0\)

\(\Leftrightarrow\frac{\left(x+3\right)\left(x-2\right)+\left(x-1\right)\left(x-4\right)+2}{\left(x-2\right)\left(x-4\right)}=0\)

\(\Leftrightarrow x^2+x-6+x^2-5x+4+2=0\)

\(\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=2\left(ktm\right)\end{cases}}\)

Vậy tập nghiệm của phương trình là \(S=\left\{0\right\}\)

ĐKXĐ: \(x\ne-2;-3;-4\)

Ta có: \(x+\frac{x}{x+2}+\frac{x+3}{x^2+5x+6}+\frac{x+4}{x^2+6x+8}=1\)

<=> \(\frac{x\left(x+2\right)}{x+2}+\frac{x}{x+2}+\frac{x+3}{\left(x+2\right)\left(x+3\right)}+\frac{x+4}{\left(x+2\right)\left(x+4\right)}\)=1

<=> \(\frac{x^2+2x}{x+2}+\frac{x}{x+2}+\frac{1}{x+2}+\frac{1}{x+2}=1\)

<=> \(\frac{x^2+3x+2}{x+2}=1\)<=>\(\frac{\left(x+1\right)\left(x+2\right)}{x+2}=1\)<=>x+1=1

<=>x=0

Vậy x=0

\(a,\frac{3}{x^2+x-2}-\frac{1}{x-1}=\frac{-7}{x+2}\left(x\ne1;x\ne-2\right)\)

\(\Leftrightarrow\frac{3}{x^2+x-2}-\frac{1}{x-1}+\frac{7}{x+2}=0\)

\(\Leftrightarrow\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{1\left(x+2\right)}{\left(x-1\right)\left(x+2\right)}+\frac{7\left(x-1\right)}{\left(x-1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{3}{\left(x-1\right)\left(x+2\right)}-\frac{x+2}{\left(x-1\right)\left(x+2\right)}+\frac{7x-7}{\left(x-1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{3-x-2+7x-7}{\left(x-1\right)\left(x+2\right)}=0\)

\(\Leftrightarrow\frac{6x-8}{\left(x-1\right)\left(x+2\right)}=0\)

=> 6x-8=0

<=> x=\(\frac{8}{6}=\frac{4}{3}\left(tmđk\right)\)

b) ĐKXĐ: x khác 2; x khác 4

\(\frac{2}{-x^2+6x-8}-\frac{x-1}{x-2}=\frac{x+3}{x-4}\)

<=> \(\frac{2}{\left(x-2\right)\left(x-4\right)}+\frac{x-1}{x-2}=\frac{x+3}{x-4}\)

<=> 2(x - 2) + (x - 1)(x - 4)(x - 2) = (x + 3)(x - 2)(x - 2)

<=> x^3 - 7x^2 + 16x - 12 = -x^3 + x^2 + 8x - 12

<=> x^2 - 7x^2 + 16x - 12 + x^3 - x^2 + 8x - 12 = 0

<=> 2x^3 - 8x^2 + 8x = 0

<=> 2x(x - 2)(x - 2) = 0

<=> 2x = 0 hoặc x - 2 = 0

<=> x = 0 (tmđk) hoặc x = 2 (ktmđk)

=> x = 2

ĐKXĐ : \(x\ne2,x\ne4\)

Phương trình ban đầu tương đương :

\(\frac{x-1}{x-2}+\frac{x+3}{x-4}+\frac{2}{x^2-6x+8}=0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)+2}{\left(x-2\right)\left(x-4\right)}=0\)

\(\Rightarrow x^2-5x+4+x^2+x-6+2=0\)

\(\Leftrightarrow2x^2-4x=0\)

\(\Leftrightarrow2x\left(x-2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-2=0\end{cases}}\)

\(\Rightarrow x=0\) ( Do x = 2 không thỏa mãn ĐKXĐ )

Vậy pt đã cho có tập nghiệm \(S=\left\{0\right\}\)

\(ĐKXĐ:x\ne2;x\ne4\)

\(\frac{x-1}{x-2}+\frac{x+3}{x-4}=\frac{2}{-x^2+6x-8}\)

\(\Rightarrow\frac{\left(x-1\right)\left(x-4\right)+\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}=\frac{-2}{x^2-6x+8}\)

\(\Rightarrow\frac{\left(x^2-5x+4\right)+\left(x^2+x-6\right)}{x^2-6x+8}=\frac{-2}{x^2-6x+8}\)

\(\Rightarrow\frac{2x^2-4x-2}{x^2-6x+8}=\frac{-2}{x^2-6x+8}\)

\(\Rightarrow2x^2-4x-2=-2\)

\(\Rightarrow2x^2-4x=0\Rightarrow2x\left(x-2\right)=0\)

\(\Rightarrow\orbr{\begin{cases}2x=0\\x-2=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\left(tm\right)\\x=2\left(ktm\right)\end{cases}}\)

Vậy pt có 1 nghiệm duy nhất là 0

\(\frac{x-1}{x-2}+\frac{x+3}{x-4}=\frac{2}{-x^2+6x-8}\)

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

\(\Leftrightarrow\frac{x-1}{x-2}+\frac{x+3}{x-4}+\frac{2}{\left(x-2\right)\left(x-4\right)}=0\)

\(\Leftrightarrow\frac{\left(x-1\right)\left(x-4\right)}{\left(x-2\right)\left(x-4\right)}+\frac{\left(x+3\right)\left(x-2\right)}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-2\right)\left(x-4\right)}=0\)

\(\Leftrightarrow\frac{x^2-5x+4}{\left(x-2\right)\left(x-4\right)}+\frac{x^2+x-6}{\left(x-2\right)\left(x-4\right)}+\frac{2}{\left(x-2\right)\left(x-4\right)}=0\)

\(\Leftrightarrow\frac{x^2-5x+4+x^2+x-6+2}{\left(x-2\right)\left(x-4\right)}=0\)

\(\Leftrightarrow2x^2+6x=0\)

\(\Leftrightarrow2x\left(x+3\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}2x=0\\x+3=0\end{cases}\Leftrightarrow\orbr{\begin{cases}x=0\\x=-3\end{cases}}}\)

Vậy x=0; x=-3

ĐK \(x\ne\left\{1;2;3;4\right\}\)

Ta có \(\frac{x^2-2x+2}{x-1}+\frac{x^2-8x+20}{x-4}=\frac{x^2-4x+6}{x-2}+\frac{x^2-6x+12}{x-3}\)

\(\Leftrightarrow\frac{\left(x-1\right)^2+1}{x-1}+\frac{\left(x-4\right)^2+4}{x-4}=\frac{\left(x-2\right)^2+2}{x-2}+\frac{\left(x-3\right)^2+3}{x-3}\)

\(\Leftrightarrow x-1+\frac{1}{x-1}+x-4+\frac{4}{x-4}=x-2+\frac{2}{x-2}+x-3+\frac{3}{x-3}\)

\(\Leftrightarrow\frac{1}{x-1}+\frac{4}{x-4}=\frac{2}{x-2}+\frac{3}{x-3}\)

\(\Leftrightarrow\frac{5x-8}{x^2-5x+4}=\frac{5x-12}{x^2-5x+6}\)\(\Leftrightarrow\left(5x-8\right)\left(x^2-5x+6\right)=\left(5x-12\right)\left(x^2-5x+4\right)\)

\(\Leftrightarrow5x^3-25x^2+30x-8x^2+40x-48=5x^3-25x^2+20x-12x^2+60x-48\)

\(\Leftrightarrow4x^2-10x=0\Leftrightarrow\orbr{\begin{cases}x=0\\x=\frac{5}{2}\end{cases}\left(tm\right)}\)

Vậy x=0 hoặc x=5/2

ĐKXĐ: \(x\ne-1,-2,-3,-4\)

\(\Leftrightarrow\frac{\left(x+1\right)^2+1}{x+1}+\frac{\left(x+4\right)^2+4}{x+4}=\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+3\right)^2+3}{x+3}\)

\(\Leftrightarrow x+1+\frac{1}{x+1}+x+4+\frac{4}{x+4}=x+2+\frac{2}{x+2}+x+3+\frac{3}{x+3}\)

\(\Leftrightarrow\frac{1}{x+1}+\frac{1}{x+4}=\frac{1}{x+2}+\frac{1}{x+3}\)

\(\Leftrightarrow\frac{x}{x+1}+\frac{x}{x+4}=\frac{x}{x+2}+\frac{x}{x+3}\)

\(\Leftrightarrow x\left(\frac{1}{x+1}+\frac{1}{x+4}-\frac{1}{x+2}-\frac{1}{x+3}\right)=0\)

\(\Leftrightarrow x\left(\frac{1}{x^2+3x+2}-\frac{1}{x^2+7x+12}\right)=0\)

\(\Leftrightarrow-x\left(\frac{4x+10}{\left(x^2+3x+2\right)\left(x^2+7x+12\right)}\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x=0\\x=-\frac{5}{2}\end{cases}}\)Thỏa mãn ĐKXĐ

Ta có Pt 

<=>\(\frac{\left(x+1\right)^2+1}{x+1}+\frac{\left(x+4\right)^2+4}{x+4}=\frac{\left(x+2\right)^2+2}{x+2}+\frac{\left(x+3\right)^2+3}{x+3}\)

<=>\(x+1+\frac{1}{x+1}+x+4+\frac{4}{x+4}=x+2+\frac{2}{x+2}+x+3+\frac{3}{x+3}\)

<=>\(\frac{1}{x+1}+\frac{4}{x+4}=\frac{2}{x+2}+\frac{3}{x+3}\)

<=>\(1-\frac{1}{x+1}+1-\frac{4}{x+4}=1-\frac{2}{x+2}+1-\frac{3}{x+3}\)

<=>\(\frac{x}{x+1}+\frac{x}{x+4}=\frac{x}{x+2}+\frac{x}{x+3}\Leftrightarrow x\left(\frac{1}{x+1}+\frac{1}{x+4}-\frac{1}{x+2}-\frac{1}{x+3}\right)=0\)

<=>\(\orbr{\begin{cases}x=0\\\frac{1}{x+1}+\frac{1}{x+4}-\frac{1}{x+2}-\frac{1}{x+3}=0\left(1\right)\end{cases}}\)

Giải pt (1) , ta có 

\(\frac{x+2-x-1}{\left(x+1\right)\left(x+2\right)}-\frac{x+4-x-3}{\left(x+3\right)\left(x+4\right)}=0\)

<=>\(\frac{1}{x^2+3x+2}-\frac{1}{x^2+7x+12}=0\Leftrightarrow x^2+3x+2=x^2+7x+12\)

<=>\(4x+10=0\Leftrightarrow x=-\frac{5}{2}\)

nhớ đối chiếu đk nhé !

^_^

de no sai sai mot chut