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a) Sửa đề: \(2x^2\left(ax^2+2bx+4c\right)=6x^4-20x^3-8x^2\)
<=> \(2ax^4+4bx^3+8cx^2=6x^4-20x^3-8x^2\)
=> \(\left\{{}\begin{matrix}2a=6\\4b=-20\\8c=-8\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=3\\b=-5\\c=-1\end{matrix}\right.\)
b) Ta có: \(\left(ax+b\right)\left(x^2-cx+2\right)=x^3+x^2-2\)
<=> \(ax^3-acx^2+2ax+bx^2-bcx+2b=x^3+x^2+2\)
<=> \(ax^3+x^2\left(b-ac\right)+x\left(2a-bc\right)+2b=x^3+x^2-2\)
=> \(\left\{{}\begin{matrix}ax^3=x^3\\\left(b-ac\right)x^2=x^2\\\left(2a-bc\right)x=0\\2b=-2\end{matrix}\right.\) <=> \(\left\{{}\begin{matrix}a=1\\b-ac=1\\2a-bc=0\\b=-1\end{matrix}\right.\)
=> a,b,c ko có!
P/s: Đề có sai ko! 
a/ \(-4x^3\cdot\left(ax^2+bx+c\right)=-8x^5+12x^4-20x^3\)
\(\Leftrightarrow-4ax^5-4bx^4-4cx^3=-8x^5+12x^4-20x^3\)
\(\Leftrightarrow\left[{}\begin{matrix}a=-\dfrac{8}{-4}=\dfrac{8}{4}=2\\b=-\dfrac{12}{4}=-3\\c=-\dfrac{20}{-4}=5\end{matrix}\right.\)
Vậy......................
b/ \(-2x^3\cdot\left(ax^2-bx-c\right)=-4x^5+6x^4+2x^3\)
\(\Leftrightarrow-2ax^5+2bx^4+2cx^3=-4x^5+6x^4+2x^3\)
\(\Leftrightarrow\left[{}\begin{matrix}a=2\\b=3\\c=1\end{matrix}\right.\)
Bài 3:
a) Ta có: \(x^3-16x=0\)
\(\Leftrightarrow x\left(x^2-16\right)=0\)
\(\Leftrightarrow x\left(x-4\right)\left(x+4\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-4=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=4\\x=-4\end{matrix}\right.\)
Vậy: \(x\in\left\{0;4;-4\right\}\)
b) Ta có: \(x^4-2x^3+10x^2-20x=0\)
\(\Leftrightarrow x\left(x^3-2x^2+10x-20\right)=0\)
\(\Leftrightarrow x\left[x^2\left(x-2\right)+10\left(x-2\right)\right]=0\)
\(\Leftrightarrow x\left(x-2\right)\left(x^2+10\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{0;2\right\}\)
c) Ta có: \(\left(2x-3\right)^2=\left(x+5\right)^2\)
\(\Leftrightarrow\left(2x-3\right)^2-\left(x+5\right)^2=0\)
\(\Leftrightarrow\left(2x-3-x-5\right)\left(2x-3+x+5\right)=0\)
\(\Leftrightarrow\left(x-8\right)\left(3x+2\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-8=0\\3x+2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\3x=-2\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-\frac{2}{3}\end{matrix}\right.\)
Vậy: \(x\in\left\{8;-\frac{2}{3}\right\}\)
d) Ta có: \(x^2\left(x-1\right)-4x^2+8x-4=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x^2-2x+1\right)=0\)
\(\Leftrightarrow x^2\left(x-1\right)-4\left(x-1\right)^2=0\)
\(\Leftrightarrow\left(x-1\right)\left[x^2-4\left(x-1\right)\right]=0\)
\(\Leftrightarrow\left(x-1\right)\left(x^2-4x+4\right)=0\)
\(\Leftrightarrow\left(x-1\right)\left(x-2\right)^2=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x-1=0\\x-2=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=1\\x=2\end{matrix}\right.\)
Vậy: \(x\in\left\{1;2\right\}\)
b: Đặt \(x^2-6x-2=a\)
Theo đề, ta có: \(a+\dfrac{14}{a+9}=0\)
=>(a+2)(a+7)=0
\(\Leftrightarrow\left(x^2-6x\right)\left(x^2-6x+5\right)=0\)
=>x(x-6)(x-1)(x-5)=0
hay \(x\in\left\{0;1;6;5\right\}\)
c: \(\Leftrightarrow\dfrac{-8x^2}{3\left(2x-1\right)\left(2x+1\right)}=\dfrac{2x}{3\left(2x-1\right)}-\dfrac{8x+1}{4\left(2x+1\right)}\)
\(\Leftrightarrow-32x^2=8x\left(2x+1\right)-3\left(8x+1\right)\left(2x-1\right)\)
\(\Leftrightarrow-32x^2=16x^2+8x-3\left(16x^2-8x+2x-1\right)\)
\(\Leftrightarrow-48x^2=8x-48x^2+18x+3\)
=>26x=-3
hay x=-3/26
\(2x^2\left(ax^2+2bx+4c\right)=6x^4-20x^3-8x^2\)
\(ax^2+2bx+4c=3x^2-10x-4\)
\(\left(a-3\right)x^2+\left(b-5\right)2x+4\left(c-1\right)=0\)
\(\left\{{}\begin{matrix}a-3=0\\b-5=0\\c-1=0\end{matrix}\right.\Rightarrow\left\{{}\begin{matrix}a=3\\b=5\\c=1\end{matrix}\right.\)
ko hiểu bạn có thể giải kĩ ra dc ko?