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a) \(=x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
\(=\left(x-1\right)^2\left(x^2+x+1\right)\)
b) \(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
c) Đổi đề: \(a^2x+a^2y-7x-7y\)
\(=a^2\left(x+y\right)-7\left(x+y\right)=\left(x+y\right)\left(a^2-7\right)\)
d) \(=x^2\left(a-b\right)+y\left(a-b\right)=\left(a-b\right)\left(x^2+y\right)\)
e) \(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
\(=\left(x+1\right)^2\left(x^2-x+1\right)\)
g) \(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h) \(=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)
i) \(=\left(x+1\right)^2-4=\left(x+1-2\right)\left(x+1+2\right)=\left(x-1\right)\left(x+3\right)\)
a\(x^3\left(x-1\right)-\left(x-1\right)=\left(x-1\right)\left(x^3-1\right)\)
b)\(=xy\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(xy-1\right)\)
d)\(=a\left(x^2+y\right)-b\left(x^2+y\right)=\left(x^2+y\right)\left(x-b\right)\)
e)\(=x^3\left(x+1\right)+\left(x+1\right)=\left(x+1\right)\left(x^3+1\right)\)
g)\(=\left(x-y\right)^2-z\left(x-y\right)=\left(x-y\right)\left(x-y-z\right)\)
h)\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)=\left(x-y\right)\left(x+y-1\right)\)
i)\(=\left(x-1\right)^2-4=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\)
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! 