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c) Ta có: \(P=x^3+y^3+6xy\)
\(=\left(x+y\right)^3-3xy\left(x+y\right)+6xy\)
\(=\left(x+y\right)^3-3xy\left(x+y-2\right)\)
\(=2^3=8\)
a) \(3xy-6xy^2=3xy\left(1-2y\right)\)
b) \(3x^3+6x^2+3x=3x\left(x^2+2x+1\right)=3x\left(x+1\right)^2\)
c) \(x^3-x^2+2\)
d) \(x^2+4x+4-y^2=\left(x^2+4x+4\right)-y^2=\left(x+2\right)^2-y^2=\left(x-y+2\right)\left(x+y+2\right)\)
e) \(x^3+4x^2+4x=x\left(x^2+4x+4\right)=x\left(x+2\right)^2\)
f) \(x^2+2x+1-9y^2=\left(x+1\right)^2-\left(3y\right)^2=\left(x-3y+1\right)\left(x+3y+1\right)\)
g) \(6x^2-12x=6x\left(x-2\right)\)
h) \(x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
i) \(x^2-2xy+y^2-9=\left(x-y\right)^2-3^2=\left(x-y-3\right)\left(x-y+3\right)\)
Bài 3 :
a )
\(4x^2-4x=0\)
\(\Leftrightarrow4x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-1=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy \(x=0\) or \(x=1\)
b )
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1.
a.\(4x^2\left(5x^3-3x+1\right)\)
\(=20x^5-12x^3+4x^2\)
b.\(\left(5x^2-4x\right)\left(x-2\right)\)
\(=5x^3-10x^2-4x^2+8x\)
\(=5x^3-14x^2+8x\)
c.\(\left(x^2-2xy+y^2\right)\left(x-y\right)\)
\(=\left(x-y\right)^2\left(x-y\right)\)
\(=\left(x-y\right)^3\)
2.
a.Bạn xem lại đề câu này nhé!
b.\(x^2-y^2-3x-3y\)
\(=\left(x^2-y^2\right)+\left(-3x-3y\right)\)
\(=\left(x+y\right)\left(x-y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-3\right)\)
3.
a.\(4x^2-4x=0\)
\(4x\left(x-1\right)=0\)
\(\Leftrightarrow\left[{}\begin{matrix}4x=0\\x-1=0\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)
Vậy x=0 hoặc x=1.
\(a,=2x^2-\dfrac{3}{2}y+3x\)
\(b,\)bt để chia hết cho x+2 là:\(2x^3+x^2-x+10\)
\(\Rightarrow m=12\)