a) \(2x^2-2x-x^2+6=0\) 

\(\Leftrightarrow x^2-2x+1+5=0\)

\(\Leftrightarrow\left(x-1\right)^2=-5\) ( vô lý)

Vậy không có x thoả mãn \(2x.\left(x-1\right)-x^2+6=0\)

b) \(x^4-2x^2.\left(3+2x^2\right)+3x^2.\left(x^2+1\right)=-3\) 

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

\(\Leftrightarrow3-3x^2=0\)

\(\Leftrightarrow3x^2=3\Leftrightarrow x^2=1\) \(\Leftrightarrow x\in\left\{-1;1\right\}\)

Vậy \(x\in\left\{-1;1\right\}\)

c) \(\left(x+1\right).\left(x^2-x+1\right)-2x=x.\left(x-2\right).\left(x+2\right)\)

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

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

\(\Leftrightarrow1+2x=0\Leftrightarrow x=\dfrac{-1}{2}\)

Vậy x=\(\dfrac{-1}{2}\)

d) \(\left(x+3\right).\left(x^2-3x+9\right)-x.\left(x-2\right).\left(x+2\right)=15\)

\(\Leftrightarrow x^3+27-x.\left(x^2-4\right)-15=0\)

\(\Leftrightarrow x^3-27-x^3+4x-15=0\)

\(\Leftrightarrow4x-42=0\)

\(\Leftrightarrow x=10,5\)

Vậy x=10,5

\(2,=\left(x-y\right)^2-2\left(x-y\right)=\left(x-y\right)\left(x-y-2\right)\\ 3,=\left(3x-5\right)\left(x+1\right)\\ 4,sai.đề\\ 5,=\left(x-1\right)^2-y^2=\left(x-y-1\right)\left(x+y-1\right)\\ 6,=\left(x+3\right)\left(x+5\right)\)

\(a,f\left(3\right)=3.3+3=9+3=12\\ f\left(-2\right)=3.\left(-2\right)+3=3-6=-3\)

\(b,f\left(m\right)=15\\ \Leftrightarrow3m+3=15\\ \Leftrightarrow3m=12\\ \Leftrightarrow m=4\)

\(4\left(x+1\right)\left(x^2-x+1\right)-x^3+3x=15\)

\(\Leftrightarrow4x^3+4-x^3+3x-15=0\)

\(\Leftrightarrow3x^3+3x-11=0\)

Đây là phương trình vô tỉ nhé bạn, nên nghiệm rất xấu!

\(\left(x-3\right)^3-\left(x-3\right)\left(x^2+3x+9\right)+9\left(x+1\right)=15\)

⇔ \(\left(x^3-3.x^2.3+3.x.3^2-3^3\right)-\left(x^3-3^3\right)+9x+9=15\)

⇔ \(x^3-9x^2+27x-27-x^3+27+9x+9=15\)

⇔ \(36x-9x^2+9=15\)

⇔ \(9x\left(4-x\right)=6\)

\(1,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\\ \Leftrightarrow\dfrac{4\left(3x+2\right)}{24}-\dfrac{6\left(3x-2\right)}{24}-\dfrac{45}{24}=0\\ \Leftrightarrow12x+24-18x+12-45=0\\ \Leftrightarrow-6x-9=0\\ \Leftrightarrow x=-\dfrac{3}{2}\)

2, ĐKXĐ:\(x\ne\pm3\)

\(\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\\ \Leftrightarrow\dfrac{\left(x+2\right)\left(3-x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{x\left(3+x\right)}{\left(3+x\right)\left(3-x\right)}-\dfrac{8x-6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow\dfrac{-x^2+x+6-3x-x^2-8x+6}{\left(3+x\right)\left(3-x\right)}=0\\ \Leftrightarrow-2x^2-10x+12=0\\ \Leftrightarrow x^2+5x-6=0\\ \Leftrightarrow\left(x-1\right)\left(x+6\right)=0\\ \Leftrightarrow\left[{}\begin{matrix}x=1\left(tm\right)\\x=-6\left(tm\right)\end{matrix}\right.\)

\(a,\dfrac{3x+2}{6}-\dfrac{3x-2}{4}=\dfrac{15}{8}\)

\(\Leftrightarrow4\left(3x+2\right)-6\left(3x-2\right)=45\)

\(\Leftrightarrow12x+8-18x+12=45\)

\(\Leftrightarrow12x-18x=45-12-8\)

\(\Leftrightarrow-6x=25\)

\(\Leftrightarrow x=\dfrac{-25}{6}\)

Vậy \(S=\left\{\dfrac{-25}{6}\right\}\)

\(b,\dfrac{x+2}{3+x}-\dfrac{x}{3-x}=\dfrac{8x-6}{9-x^2}\left(ĐKXĐ:x\ne3;x\ne-3\right)\)

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

\(\Leftrightarrow3x-x^2+6-2x-3x-x^2=8x-6\)

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

\(\Leftrightarrow-2x^2-10x=0\)

\(\Leftrightarrow-2x\left(x-5\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}-2x=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\left(nhận\right)\\x=5\left(nhận\right)\end{matrix}\right.\)

Vậy \(S=\left\{0;5\right\}\)