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: \(-2-x^2+5x-4=x^2+x-6\)

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

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

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

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

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

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

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

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

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

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

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

\(\Rightarrow6x-6=0\)

\(\Rightarrow x=1\)

Tu phuong trinh da cho suy ra

(x-1)(x+1)(x+2)(x+4)=0

x=1;-1;-1;-4

\(x^4+6x^3+7x^2-6x+1=9\)               

\(\Leftrightarrow x^4+6x^3+7x^2-6x-8=0\)

\(\Leftrightarrow x^4+x^3+5x^3+5x^2+2x^2+2x-8x-8=0\)

\(\Leftrightarrow\left(x^4+x^3\right)+\left(5x^3+5x^2\right)+\left(2x^2+2x\right)-\left(8x+8\right)=0\)

\(\Leftrightarrow x^3\left(x+1\right)+5x^2\left(x+1\right)+2x\left(x+1\right)-8\left(x+1\right)=0\)

\(\Leftrightarrow\left(x+1\right)\left(x^3+5x^2+2x-8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+1=0\\x^3+5x^2+2x-8=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-1\\x^3+5x^2+2x-8=0\end{cases}}\)

\(x^3+5x^2+2x-8=0\)

\(\Leftrightarrow x^3-x^2+6x^2-6x+8x-8=0\)

\(\Leftrightarrow\left(x^3-x^2\right)+\left(6x^2-6x\right)+\left(8x-8\right)=0\)

\(\Leftrightarrow x^2\left(x-1\right)+6x\left(x-1\right)+8\left(x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left(x^2+6x+8\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=0\\x^2+6x+8=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x^2+4x+2x+8=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\x\left(x+4\right)+2\left(x+4\right)=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=1\\\left(x+4\right)\left(x+2\right)=0\end{cases}}\)

\(\left(x+4\right)\left(x+2\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x+4=0\\x+2=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-4\\x=-2\end{cases}}\)

Vậy \(x=\left\{-4;-2;-1;1\right\}\)

ĐKXĐ: \(\left\{{}\begin{matrix}x\ge1\\\frac{-1-\sqrt{5}}{4}\le x\le-\frac{1}{8}\end{matrix}\right.\)(Có thể chưa chính xác)

\(12x^2+16x+1=2\sqrt{24x^3+12x^2-6x}+4\sqrt{x^2-x}+4\sqrt{8x^3+9x^2+x}\)

Áp dụng AM-GM:

\(2\sqrt{24x^3+12x^2-6x}=2\sqrt{6x\left(4x^2+2x-1\right)}\le6x+\left(4x^2+2x-1\right)=4x^2+8x-1\left(1\right)\)

\(4\sqrt{x^2-x}=2\sqrt{1.\left(4x^2-4x\right)}\le4x^2-4x+1\left(2\right)\)

\(4\sqrt{8x^3+9x^2+x}=2\sqrt{\left(4x^2+4x\right)\left(8x+1\right)}\le\left(4x^2+4x\right)+\left(8x+1\right)=4x^2+12x+1\left(3\right)\)

Cộng \(\left(1\right),\left(2\right),\left(3\right)\), ta có: \(VP\le VT\)

Dấu ''='' xảy ra khi :

\(\left\{{}\begin{matrix}4x^2+2x-1=6x\\4x^2-4x=1\\4x^2+4x=8x+1\end{matrix}\right.\)\(\Rightarrow4x^2-4x-1=0\)

\(\Rightarrow x=\frac{1\pm\sqrt{2}}{2}\) (t/m ĐKXĐ)

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 \(x=-\dfrac{1}{13}\)

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:

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

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

<=> \(\frac{6\left(x+4\right)-30x+120}{30}=\frac{10x-15x+30}{30}\)

<=> 6x + 24 - 30x + 120 = -5x + 30

<=> -24x + 5x = 30 - 144

<=> -19x = -114

<=> x = 6

Vậy S = {6}

\(6x^4-x^3-7x^2+x+1=0\)

\(\Leftrightarrow\left(6x^4-6x^3\right)+\left(5x^3-5x^2\right)+\left(-2x^2+2x\right)+\left(-x+1\right)=0\)\(\Leftrightarrow\left(x-1\right)\left(6x^3+5x^2-2x-1\right)=0\)

\(\Leftrightarrow\left(x-1\right)\left[\left(6x^3-3x^2\right)+\left(8x^2-4x\right)+\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left[3x^2\left(2x-1\right)+4x\left(2x-1\right)+\left(2x-1\right)\right]=0\)

\(\Leftrightarrow\left(x-1\right)\left(2x-1\right)\left(3x^2+4x+1\right)=0\)

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

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

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

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

Ta có : \(6x^4+5x^3-38x^2+5x+6=0\)

\(\Leftrightarrow6x^4+20x^3+6x^2-15x^3-50x^2-15x+6x^2+20x+6=0\)

\(\Leftrightarrow2x^2\left(3x^2+10x+3\right)-5x\left(3x^2+10x+3\right)+2\left(3x^2+10x+3\right)=0\)

\(\Leftrightarrow\left(3x^2+10x+3\right)\left(2x^2-5x+2\right)=0\)

\(\Leftrightarrow\left(3x^2+x+9x+3\right)\left(2x^2-x-4x+2\right)=0\)

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

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

\(\Leftrightarrow\)\(3x+1=0\)

hoặc    \(x+3=0\)

hoặc   \(2x-1=0\)

hoặc    \(x-2=0\)

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

hoặc   \(x=-3\)

hoặc   \(x=\frac{1}{2}\)

hoặc   \(x=2\)

Vậy tập nghiệm của phương trình là \(S=\left\{-\frac{1}{3};-3;\frac{1}{2};2\right\}\)