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\(1,\\ a,A=4x^2\left(-3x^2+1\right)+6x^2\left(2x^2-1\right)+x^2\\ A=-12x^4+4x^2+12x^2-6x^2+x^2=-x^2=-\left(-1\right)^2=-1\\ b,B=x^2\left(-2y^3-2y^2+1\right)-2y^2\left(x^2y+x^2\right)\\ B=-2x^2y^3-2x^2y^2+x^2-2x^2y^3-2x^2y^2\\ B=-4x^2y^3-4x^2y^2+x^2\\ B=-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^3-4\left(0,5\right)^2\left(-\dfrac{1}{2}\right)^2+\left(0,5\right)^2\\ B=\dfrac{1}{8}-\dfrac{1}{4}+\dfrac{1}{4}=\dfrac{1}{8}\)
\(2,\\ a,\Leftrightarrow10x-16-12x+15=12x-16+11\\ \Leftrightarrow-14x=-4\\ \Leftrightarrow x=\dfrac{2}{7}\\ b,\Leftrightarrow12x^2-4x^3+3x^3-12x^2=8\\ \Leftrightarrow-x^3=8=-2^3\\ \Leftrightarrow x=2\\ c,\Leftrightarrow4x^2\left(4x-2\right)-x^3+8x^2=15\\ \Leftrightarrow16x^3-8x^2-x^3+8x^2=15\\ \Leftrightarrow15x^3=15\\ \Leftrightarrow x^3=1\Leftrightarrow x=1\)
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a) \(P=3\left(x^2+2xy+y^2\right)-2\left(x+y\right)-100\)
\(P=3\left(x+y\right)^2-2.5-100\)
\(P=3.5^2-110\)
\(P=-35\)
b) \(Q=\left[x^3+y^3+3xy\left(x+y\right)\right]-2\left(x^2+2xy+y^2\right)+3.5+10\)
\(Q=\left(x+y\right)^3-2\left(x+y\right)^2+25\)
\(Q=5^3-2.5^2+25\)
\(Q=100\)
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k: \(=\left(2-3x\right)\left(4+6x+9x^2\right)\)
i: \(=3\left(x^2-2xy+y^2\right)=3\left(x-y\right)^2\)
\(g,27+27x+9x^2+x^3=\left(3+x\right)^3\\ i,2x^2+2y^2-x^2z+z-y^2z-2=\left(2x^2-x^2z\right)+\left(2y^2-y^2z\right)-\left(2-z\right)=x^2\left(2-z\right)+y^2\left(2-z\right)-\left(2-z\right)=\left(x^2+y^2-1\right)\left(2-z\right)\)
\(k,8-27x^2=2^3-\left(3x\right)^3=\left(2-3x\right)\left(4+6x+9x^2\right)\)
\(l,3x^2-6xy+3y^2=3\left(x^2-2xy+y^2\right)=3\left(x-y\right)^2\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(2x^2-4xy+2y^2\\ =2\left(x^2-2xy+y^2\right)\\ =2\left(x-y\right)^2\)
a) 2x2-4xy+2y2
= 2x2-2xy-2xy+2y2
= 2x(x-y)-2y(x-y)
= (2x-2y)(x-y)
b) x2+4xy+4y2-9
= (x+2y)2-32
= (x+2y-3)(x+2y+3)
c) x4-x3y+x-y
= x3(x-y)+(x-y)
= (x3+1)(x-y)
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a.
$12x^3y-24x^2y^2+12xy^3=12xy(x^2-2xy+y^2)=12xy(x-y)^2$
b.
$x^2-6x+xy-6y=(x^2+xy)-(6x+6y)=x(x+y)-6(x+y)=(x-6)(x+y)$
c.
$2x^2+2xy-x-y=2x(x+y)-(x+y)=(x+y)(2x-1)$
d.
$x^3-3x^2+3x-1=(x-1)^3$
e.
$3x^2-3y^2-12x-12y=(3x^2-3y^2)-(12x+12y)$
$=3(x-y)(x+y)-12(x+y)=(x+y)[3(x-y)-12]=3(x-y)(x-y-4)$
f.
$x^2-2xy-x^2+4y^2=4y^2-2xy=2y(2y-x)$
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\(x^3+y^3-2x^2-2y^2+3xy\left(x+y\right)-4xy+3\left(x+y\right)+10=\left[x^3+y^3+3xy\left(x+y\right)\right]-2\left(x^2+2xy+y^2\right)+3\left(x+y\right)+10=\left(x+y\right)^3-2\left(x+y\right)^2+3\left(x+y\right)+10=5^3-2.5^2+3.5+10=100\)
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a)\(3x\left(x-1\right)+2x^2\left(x-1\right)=0\\ \Leftrightarrow x\left(x-1\right)\left(3+2x\right)=0\\ \Leftrightarrow\left\{{}\begin{matrix}x=0\\x-1=0\\3+2x=0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x=0\\x=1\\x=\dfrac{-3}{2}\end{matrix}\right.\)
a: Ta có: \(3x^2-3x+2x^3-2x^2=0\)
\(\Leftrightarrow2x^3+x^2-3x=0\)
\(\Leftrightarrow x\left(2x^2+x-3\right)=0\)
\(\Leftrightarrow x\left(x-1\right)\left(2x+3\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\\x=-\dfrac{3}{2}\end{matrix}\right.\)
b: Ta có: \(x^3+27=-x^2+9\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9\right)+\left(x+3\right)\left(x-3\right)=0\)
\(\Leftrightarrow\left(x+3\right)\left(x^2-3x+9+x-3\right)=0\)
\(\Leftrightarrow x+3=0\)
hay x=-3