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\(x^2\left(x+1\right)-\left(x+1\right)\left(3x+1\right)+7x-x^2\)
\(=x^3+x^2-3x^2-4x-1+7x-x^2\)
\(=x^3-3x^2+3x-1\)
\(=\left(x-1\right)^3\)
a) 2x + 2y - x2 - xy
= 2(x + y) + x(x + y)
= (x + y) (x + 2)
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a)\(2x+2y-x^2-xy=2\left(x+y\right)-x\left(x+y\right)=\left(2-x\right)\left(x+y\right)\)
b)\(\left(x+3\right)^2-\left(2x-5\right)\left(x+3\right)\)
\(=\left(x+3\right)\left[\left(x+3\right)-\left(2x-5\right)\right]\)
\(=\left(x+3\right)\left(8-x\right)\)
c)\(\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(9x^2-4\right)\)
\(=\left(3x+2\right)^2+\left(3x-2\right)^2-2\left(3x-2\right)^2\)
\(=\left(3x+2\right)\left[\left(3x+2\right)-\left(3x-2\right)\right]+\left(3x-2\right)\left[\left(3x-2\right)-\left(3x+2\right)\right]\)
\(=4\left(3x+2\right)-4\left(3x-2\right)\)
\(=4\left(3x+2-3x+2\right)\)
=4.4=16
\(A=3x^2-14x^2+4x+3\)
Giả sử:
\(A=\left(3x+a\right)\left(x^2+bx+c\right)\)
\(=3x^3+3bx^2+3cx+ax^{2\:}+abx+ac\)
\(=3x^3+\left(3b+a\right)x^2+\left(3c+ab\right)x+ac\)
Ta có:
\(\begin{cases}3b+a=-14\\3c+ab=4\\ac=3\end{cases}\)\(\Rightarrow\begin{cases}a=1\\b=-5\\c=3\end{cases}\)
Vậy \(A=\left(3x+1\right)\left(x^2-5x+3\right)\)
a) \(9x^2-12x+4\)
\(=9x^2-6x-6x+4\)
\(=3x\left(3x-2\right)-2\left(3x-2\right)\)
\(=\left(3x-2\right)^2\)
b) \(2xy+16-x^2-y^2\)
\(=-\left(x^2-2xy+y^2-16\right)\)
\(=-\left(x-y\right)^2+16\)
\(=\left(4-x+y\right)\left(4+x-y\right)\)
c) \(3x+2x^2-2\)
\(=2x^2+4x-x-2\)
\(=2x\left(x+2\right)-\left(x+2\right)=\left(x+2\right)\left(2x-1\right)\)