a: Ta có: \(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=\dfrac{-7}{x+2}\)

\(\Leftrightarrow3-\left(x+2\right)=-7\left(x-1\right)\)

\(\Leftrightarrow3-x-2+7x-7=0\)

\(\Leftrightarrow6x-6=0\)

hay x=1(loại

b: Ta có: \(\dfrac{2}{-x^2+6x-8}-\dfrac{x-1}{x-2}=\dfrac{x+3}{x-4}\)

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

Suy ra: \(-2-x^2+5x-4=x^2+x-6\)

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

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

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

\(\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=2\left(loại\right)\end{matrix}\right.\)

\(\dfrac{3}{x^2+x-2}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

\(\Rightarrow\dfrac{3}{\left(x^2-x\right)+\left(2x-2\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

\(\Rightarrow\dfrac{3}{x\left(x-1\right)+2\left(x-1\right)}-\dfrac{1}{x-1}=-\dfrac{7}{x+2}\)

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

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

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

\(\Rightarrow3-x-2+7x-7=0\)

\(\Rightarrow6x-6=0\)

\(\Rightarrow x=1\)

         (12x-1)(6x-1)(4x-1)(3x-1)=5

<=>(12x-1)(12x-2)(12x-3)(12x-4)=40

<=>[(12x-1)(12x-4)] [(12x-2)(12x-3)] =40

<=>(144x^2 - 60x + 4) (144x^2 - 60x + 6) =40

đặt 144x^2 - 60x +4 = t  =>144x^2 - 60x +6 = t+2

ta có phương trình:

        t ( t+2 ) =40

<=> t^2 + 2t -40 =0

<=> (t+1)^2 -39 =0

<=> t+1=\(\sqrt{39}\)      hoặc t+1=\(-\sqrt{39}\)    <=> x=\(\sqrt{39}\) -1 hoặc x=\(-\sqrt{39}\) -1

tick nha

b: Ta có: \(\dfrac{x+1}{x-2}-\dfrac{x-1}{x+2}=\dfrac{2\left(x^2+2\right)}{x^2-4}\)

\(\Leftrightarrow x^2+3x+2-x^2+3x-2-2x^2-4=0\)

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

a=-2; b=6; c=-4

Vì a+b+c=0 nên phương trình có hai nghiệm phân biệt là:

\(x_1=1\left(nhận\right);x_2=\dfrac{c}{a}=2\left(loại\right)\)

\(\dfrac{180}{x-4}-\dfrac{180}{x}=\dfrac{1}{2}\)

\(\Leftrightarrow\) \(\dfrac{2x\cdot180}{2x\left(x-4\right)}-\dfrac{2\cdot180\cdot\left(x-4\right)}{2x\left(x-4\right)}=0\)

\(\Leftrightarrow\) \(\dfrac{360x-360x+1440-x^2+4x}{2x\left(x-4\right)}=0\)

\(\Leftrightarrow\) \(\dfrac{-x^2+4x+1440}{2x\left(x-4\right)}=0\)

\(\Leftrightarrow-x^2+4x+1440=0\)

\(\Leftrightarrow-x^2+40x-36x+1440=0\)

\(\Leftrightarrow-x\cdot\left(x-40\right)\cdot\left(-36\right)\cdot\left(x-40\right)=0\)

\(\Leftrightarrow\left(x-40\right)\cdot\left(x-36\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x-40=0\\x+36=0\end{matrix}\right.\)

 \(x-40=0\)

  \(x=0+40\)

 \(x=40\)

\(x+36=0\)

   \(x=0-36\)

   \(x=-36\)

\(\Leftrightarrow\left[{}\begin{matrix}x=40\\x=-36\end{matrix}\right.\)

\(180\left(\dfrac{1}{x-4}-\dfrac{1}{x}\right)=\dfrac{1}{2}\)

\(\dfrac{1}{x-4}-\dfrac{1}{x}=\dfrac{1}{360}\left(đk:x\ne0,4\right)\)

\(\dfrac{x-x+4}{x\left(x-4\right)}=\dfrac{1}{360}\)

\(\dfrac{4}{x\left(x-4\right)}=\dfrac{1}{360}\)

\(x^2-4x=1440\)

\(x^2-4x+4=1444\)

\(\left(x-2\right)^2=1444=38^2\)

\(\Leftrightarrow\left[{}\begin{matrix}x-2=38\\x-2=-38\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=40\\x=-36\end{matrix}\right.\)

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

\(\Leftrightarrow\dfrac{\left(2x-5\right)\left(x-1\right)-2x\left(x+3\right)}{\left(x-1\right)\left(x+3\right)}=\dfrac{4}{\left(x+3\right)\left(x-1\right)}\)

Suy ra: \(2x^2-2x-5x+5-2x^2-6x=4\)

\(\Leftrightarrow13x=-1\)

hay \(x=-\dfrac{1}{13}\)

\(ĐK:x\ne\pm\dfrac{3}{2}\\ PT\Leftrightarrow2x+3+2x-3=2x+4\\ \Leftrightarrow2x=4\Leftrightarrow x=2\left(tm\right)\)

\(\dfrac{1}{2x-3}+\dfrac{1}{2x+3}=\dfrac{2x+4}{4x^2-9}\)

\(\dfrac{2x+3+2x-3}{\left(2x-3\right)\left(2x+3\right)}=\dfrac{2x+4}{4x^2-9}\)

\(\dfrac{4x}{4x^2-9}=\dfrac{2x+4}{4x^2-9}\Rightarrow4x=2x+4\)

\(\Rightarrow2x=4\Rightarrow x=2\)

  \(x\left(x-1\right)\left(x+1\right)\left(x+2\right)=24\)

\(\Leftrightarrow\left[x\left(x+1\right)\right]\left[\left(x-1\right)\left(x+2\right)\right]=24\)

\(\Leftrightarrow\left(x^2+x\right)\left(x^2+x-2\right)=24\)

Đặt \(x^2+x-1=a\)

Ta có : \(x^2+x-1=\left(x+\frac{1}{2}\right)^2-\frac{5}{4}\ge-\frac{5}{4}\)

\(\Rightarrow a\ge-\frac{5}{4}\)

Ta có pt : \(\left(a+1\right)\left(a-1\right)=24\)

\(\Leftrightarrow a^2-1=24\)

\(\Leftrightarrow a^2=25\)

\(\Leftrightarrow a=5\left(Do\text{ }a\ge-\frac{5}{4}\right)\)

\(\Leftrightarrow x^2+x-1=5\)

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

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

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

a, \(\dfrac{6-x}{4x-3}=\dfrac{2}{4x-3}\)

ĐKXĐ: \(x\ne\dfrac{3}{4}\)

PT đã cho \(\Leftrightarrow\)\(\dfrac{\left(6-x\right)\left(4x-3\right)}{4x-3}=\dfrac{2\left(4x-3\right)}{4x-3}\)

                  \(\Rightarrow6-x=2\)

                  \(\Leftrightarrow x=4\)(thỏa mãn ĐKXĐ)

b, \(\dfrac{3-x}{2x-3}+x-1=\dfrac{-4}{2x-3}\)

ĐKXĐ: \(x\ne\dfrac{3}{2}\)

PT đã cho \(\Leftrightarrow\)\(\dfrac{\left(3-x\right)\left(2x-3\right)}{2x-3}+\left(x+1\right)\left(2x-3\right)=\dfrac{-4\left(2x-3\right)}{2x-3}\)

                  \(\Rightarrow3-x+2x-3x+2x-3=-8x+12\)

                  \(\Leftrightarrow8x=12\)

                  \(\Leftrightarrow x=\dfrac{3}{2}\)(không thỏa mãn ĐKXĐ)

Vậy \(x\in\varnothing\).

a) ĐK: \(x\ne\dfrac{3}{4}\)

PT \(\Rightarrow27x-18-4x^2=8x-6\)

\(\Leftrightarrow4x^2-19x+12=0\) \(\Leftrightarrow\left[{}\begin{matrix}x=4\left(nhận\right)\\x=\dfrac{3}{4}\left(loại\right)\end{matrix}\right.\)

  Vậy phương trình có nghiệm \(x=4\)

b) ĐK: \(x\ne\dfrac{3}{2}\)

PT \(\Rightarrow3-x+2x^2-5x+3=-4\) 

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

  Vậy phương trình vô nghiệm

c) ĐK: \(x\ne3\)

PT \(\Rightarrow2x^2-5x-3=2x-4\)

\(\Leftrightarrow2x^2-7x+1=0\) \(\Leftrightarrow x=\dfrac{7\pm\sqrt{41}}{4}\)

  Vậy phương trình có nghiệm \(x=\dfrac{7\pm\sqrt{41}}{4}\)