\(ĐK:x\ne3;x\ne2\\ PT\Leftrightarrow\dfrac{x^2+3x+2}{x-3}\left(\dfrac{x+1}{x-2}+1+\dfrac{x^2}{x-2}\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{\left(x+1\right)\left(x+2\right)}{x-3}=0\\\dfrac{x^2+x+2}{x-2}=0\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x^2+x+2=0\left(vô.n_0\right)\end{matrix}\right.\\ \Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\end{matrix}\right.\)

ĐKXĐ: \(x\notin\left\{2;-1;\dfrac{-3\pm\sqrt{17}}{2}\right\}\)

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

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

=>\(\dfrac{x^3+3x^2-2x+3x^3-3x^2-6x}{\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2}=1\)

=>\(4x^3-8x=\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2\)

=>\(4x\left(x^2-2\right)=\left(x^2-2\right)^2+2x\left(x^2-2\right)-3x^2\)

=>\(\left(x^2-2\right)^2-2x\left(x^2-2\right)-3x^2=0\)

=>\(\left(x^2-2\right)^2-3x\left(x^2-2\right)+x\left(x^2-2\right)-3x^2=0\)

=>\(\left(x^2-2\right)\left(x^2-2-3x\right)+x\left(x^2-2-3x\right)=0\)

=>\(\left(x^2+x-2\right)\left(x^2-3x-2\right)=0\)

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

=>\(\left[{}\begin{matrix}x+2=0\\x-1=0\\x^2-3x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-2\left(nhận\right)\\x=1\left(nhận\right)\\x=\dfrac{3\pm\sqrt{17}}{2}\left(nhận\right)\end{matrix}\right.\)

PT 2 

\(\Leftrightarrow\dfrac{3}{\left(x-1\right)\left(x-2\right)\left(x-3\right)}+\dfrac{2x}{\left(x-2\right)\left(x-3\right)}-\dfrac{1}{\left(x-1\right)\left(x-2\right)}=0\) ( \(x\ne1;x\ne2;x\ne3\))

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

\(\Rightarrow2x^2-3x+6=0\)

=> PT vô nghiệm.

ĐKXĐ : \(x\inℝ\)

Ta có : \(\dfrac{x^2+4x+5}{x^2-x+5}-\dfrac{3x}{x^2-3x+5}=1\)

\(\Leftrightarrow1+\dfrac{5x}{x^2-x+5}-\dfrac{3x}{x^2-3x+5}=1\)

\(\Leftrightarrow x.\left(\dfrac{5}{x^2-x+5}-\dfrac{3}{x^2-3x+5}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\\dfrac{5}{x^2-x+5}=\dfrac{3}{x^2-3x+5}\left(1\right)\end{matrix}\right.\)

Phương trình (1) <=> 5(x² - 3x + 5) = 3(x² - x + 5)

<=> 2x² - 12x + 10 = 0

<=> x² - 6x + 5 = 0

<=> (x - 1)(x - 5) = 0

<=> \(\left[{}\begin{matrix}x=1\\x=5\end{matrix}\right.\)

Tập nghiệm \(S=\left\{0;1;5\right\}\)

\(ĐK:x\ne3\\ PT\Leftrightarrow\dfrac{x^2+3x+2}{x-3}\left(-x-1+x^2-2x-7\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}\dfrac{\left(x+1\right)\left(x+2\right)}{x-3}=0\\x^2-3x-8=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-1\\x=-2\\x=\dfrac{3+\sqrt{41}}{2}\\x=\dfrac{3-\sqrt{41}}{2}\end{matrix}\right.\)

Đk: \(x\ne5;x\ne-10\)

Pt: \(\Rightarrow\dfrac{\left(x-2\right)\left(x+5\right)}{x^2}-\dfrac{40}{\left(x-5\right)\left(x+10\right)}=0\)

     \(\Rightarrow\left(x-2\right)\left(x+5\right)\left(x-5\right)\left(x+10\right)-40x^2=0\)

     \(\Rightarrow\left(x^2-12x+20\right)\left(x^2-25\right)-40x^2=0\)

     \(\Rightarrow x^4-12x^3-45x^2+300x=500\)

     \(\Rightarrow\left\{{}\begin{matrix}x=5\left(loại\right)\\x=-5\left(tm\right)\end{matrix}\right.\)

`3/(x^2-3x+3)+x^2-3x-3=0`

`<=>3+(x^2-3x-3)(x^2-3x+3)=0`

`<=>3+(x^2-3x)^2-9=0`

`<=>(x^2-3x)^2-6=0`

`<=>x^2-3x=+-6`

Đến đây chia 2 th rồi giải thôi :v

Căn 6 nhé

Đặt \(\left\{{}\begin{matrix}\dfrac{x}{x-1}=a\\\dfrac{1}{y+2}=b\end{matrix}\right.\)

\(\Rightarrow\)Ta có hệ mới: \(\left\{{}\begin{matrix}3a-2b=4\\2a+b=5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}2\cdot\left(3a-2b\right)=2\cdot4\\3\left(2a+b\right)=3\cdot5\end{matrix}\right.\)

\(\Rightarrow\left\{{}\begin{matrix}6a-4b=8\left(1\right)\\6a+3b=15 \left(2\right)\end{matrix}\right.\)

Lấy (2)-(1) ta đc:

\(\Rightarrow7b=7\Rightarrow b=1\Rightarrow2a+1=5\Rightarrow a=2\)

\(\Rightarrow\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\\dfrac{1}{y+2}=1\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\left(x-1\right)\\1=y+2\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}x=2\\y=-1\end{matrix}\right.\)

Với \(x\ne1;y\ne-2\)

hpt <=>\(\left\{{}\begin{matrix}\dfrac{3x}{x-1}-\dfrac{2}{y+2}=4\\\dfrac{4x}{x-1}+\dfrac{2}{y+2}=10\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}\dfrac{7x}{x-1}=14\\\dfrac{2x}{x-1}+\dfrac{1}{y+2}=5\end{matrix}\right.\)

<=>\(\left\{{}\begin{matrix}\dfrac{x}{x-1}=2\\2.2+\dfrac{1}{y+2}=5\end{matrix}\right.\)

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

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

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