Phân tích các đa thức sau thành nhân tử. Lời giải+ đáp án
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\(a^2-4b^2=a^2-\left(2b\right)^2=\left(a-2b\right)\left(a+2b\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3x^2-3y^2-2\left(x^2-2xy+y^2\right)\)
\(=3x^2-3y^2-2x^2+4xy-2y^2\)
\(=x^2+4xy-5y^2\)
\(=x^2+4xy+4y^2-9y^2\)
\(=\left(x+2y\right)^2-\left(3y\right)^2\)
\(=\left(x+2y-3y\right)\left(x+2y+3y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(x^2-4x-y^2+4=\left(x^2-4x+4\right)-y^2=\left(x-2\right)^2-y^2=\left(x-2-y\right)\left(x-2+y\right)\)
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\(x^2-2x-3=x^2-3x+x-3=x\left(x-3\right)+\left(x-3\right)=\left(x+1\right)\left(x-3\right)\)
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a) x3-2x2-x+2
=x(x2-1)+2(-x2+1)
=x(x2-1)-2(x2-1)
=(x2-1)(x-2)
b)
x2+6x-y2+9
=x2+6x+9-y2
=(x+3)2-y2
=(x+3-y)(x+3+y)
![](https://rs.olm.vn/images/avt/0.png?1311)
d: \(9x^2-25=\left(3x-5\right)\left(3x+5\right)\)
e: \(16x^2-4y^2=4\left(4x^2-y^2\right)=4\left(2x-y\right)\left(2x+y\right)\)
f: \(\left(x+\dfrac{1}{4}\right)^2=x^2+\dfrac{1}{2}x+\dfrac{1}{16}\)
\(d,9x^2-25=\left(3x-5\right)\left(3x+5\right)\\ e,16x^2-4y^2=\left(4x-2y\right)\left(4x+2y\right)=4\left(2x-y\right)\left(2x+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
Xóa
\(5x\left(x-2\right)-x+2=0\)
\(\Rightarrow5x\left(x-2\right)-\left(x-2\right)=0\)
\(\Rightarrow\left(5x-1\right)\left(x-2\right)=0\)
\(\Rightarrow\left[{}\begin{matrix}5x-1=0\\x-2=0\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{1}{5}\\x=2\end{matrix}\right.\)
Vậy x=1/5 hoặc x=2