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tính:
a) \(3x^2\left(2x-1\right)\)
phân tích đa thức thành nhân tử:
a) \(5x^2-10x\)
b) \(4x^2-y^2-4x+1\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(=5x\left(x-2\right)\)
b) \(=\left(2x\right)^2-2x.2+1-y^2=\left(2x-1\right)^2-y^2=\left(2x-1-y\right)\left(2x-1+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a, x^3 + 4x = x(x^2+4)`
`b, 6ab - 9ab^2 = 3ab(2-b)`
`c, 2a(x-1) + 3b(1-x)`
`= (2a-3b)(x-1)`
`d, (x-y)^2 - x(y-x)`
`= (x-y+x)(x-y)`
`= (2x-y)(x-y)`
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(4x^2-1=\left(2x+1\right)\left(2x-1\right)\)
b) \(\left(x+2\right)^2-9=\left(x-1\right)\left(x+5\right)\)
c) \(\left(a+b\right)^2-\left(a-2b\right)^2\)
\(=\left(a+b-a+2b\right)\left(a+b+a-2b\right)\)
\(=3b\left(2a-b\right)\)
`a, 4x^2-1 = (2x+1)(2x-1)`
`b, (x+2)^2-9 = (x+2-3)(x+2+3) = (x-1)(x+5)`
`c, (a+b)^2-(a-2b)^2 = (a+b+a-2b)(a+b-a+2b) = (2a-b)(3b)`
![](https://rs.olm.vn/images/avt/0.png?1311)
a: \(x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2+3\left(x-y\right)-4\)
\(=\left(x-y+4\right)\left(x-y-1\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
`a, 4a^2 + 4a + 1 = (2a+1)^2`
`b, -3x^2 + 6xy - 3y^2`
` = -3(x-y)^2`
`c, (x+y)^2 - 2(x+y)z + z^2`
`= (x+y-z)^2`
![](https://rs.olm.vn/images/avt/0.png?1311)
Bài 2:
1: \(\left(2x-1\right)^2-4\left(2x-1\right)=0\)
=>\(\left(2x-1\right)\left(2x-1-4\right)=0\)
=>(2x-1)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-1=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{1}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
2: \(9x^3-x=0\)
=>\(x\left(9x^2-1\right)=0\)
=>x(3x-1)(3x+1)=0
=>\(\left[{}\begin{matrix}x=0\\3x-1=0\\3x+1=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{3}\\x=-\dfrac{1}{3}\end{matrix}\right.\)
3: \(\left(3-2x\right)^2-2\left(2x-3\right)=0\)
=>\(\left(2x-3\right)^2-2\left(2x-3\right)=0\)
=>(2x-3)(2x-3-2)=0
=>(2x-3)(2x-5)=0
=>\(\left[{}\begin{matrix}2x-3=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=\dfrac{3}{2}\\x=\dfrac{5}{2}\end{matrix}\right.\)
4: \(\left(2x-5\right)\left(x+5\right)-10x+25=0\)
=>\(2x^2+10x-5x-25-10x+25=0\)
=>\(2x^2-5x=0\)
=>\(x\left(2x-5\right)=0\)
=>\(\left[{}\begin{matrix}x=0\\2x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{5}{2}\end{matrix}\right.\)
Bài 1:
1: \(3x^3y^2-6xy\)
\(=3xy\cdot x^2y-3xy\cdot2\)
\(=3xy\left(x^2y-2\right)\)
2: \(\left(x-2y\right)\left(x+3y\right)-2\left(x-2y\right)\)
\(=\left(x-2y\right)\cdot\left(x+3y\right)-2\cdot\left(x-2y\right)\)
\(=\left(x-2y\right)\left(x+3y-2\right)\)
3: \(\left(3x-1\right)\left(x-2y\right)-5x\left(2y-x\right)\)
\(=\left(3x-1\right)\left(x-2y\right)+5x\left(x-2y\right)\)
\(=(x-2y)(3x-1+5x)\)
\(=\left(x-2y\right)\left(8x-1\right)\)
4: \(x^2-y^2-6y-9\)
\(=x^2-\left(y^2+6y+9\right)\)
\(=x^2-\left(y+3\right)^2\)
\(=\left(x-y-3\right)\left(x+y+3\right)\)
5: \(\left(3x-y\right)^2-4y^2\)
\(=\left(3x-y\right)^2-\left(2y\right)^2\)
\(=\left(3x-y-2y\right)\left(3x-y+2y\right)\)
\(=\left(3x-3y\right)\left(3x+y\right)\)
\(=3\left(x-y\right)\left(3x+y\right)\)
6: \(4x^2-9y^2-4x+1\)
\(=\left(4x^2-4x+1\right)-9y^2\)
\(=\left(2x-1\right)^2-\left(3y\right)^2\)
\(=\left(2x-1-3y\right)\left(2x-1+3y\right)\)
8: \(x^2y-xy^2-2x+2y\)
\(=xy\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(xy-2\right)\)
9: \(x^2-y^2-2x+2y\)
\(=\left(x^2-y^2\right)-\left(2x-2y\right)\)
\(=\left(x-y\right)\left(x+y\right)-2\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(a,\left(x-1\right)^2-2^2=\left(x-1-2\right)\left(x-1+2\right)=\left(x-3\right)\left(x+1\right)\\ b,=\left(2x\right)^2+2.2x.3+3^2\\ =\left(2x+3\right)^2\\ c,=x^3-\left(2y\right)^3\\ =\left(x-2y\right)\left(x^2+2xy+4y^2\right)\\ d,=x^3\left(x^2-1\right)-\left(x^2-1\right)\\ =\left(x^3-1\right)\left(x^2-1\right)\\ =\left(x-1\right)\left(x^2+x+1\right)\left(x-1\right)\left(x+1\right)\\ =\left(x-1\right)\left(x+1\right)\left(x^2+x+1\right)\)
\(e,=-4x^2\left(x-1\right)+\left(x-1\right)\\ =\left(1-4x^2\right)\left(x-1\right)\\ =\left(1-2x\right)\left(1+2x\right)\left(x-1\right)\)
\(f,=\left(2x\right)^3+3.\left(2x\right)^2.1+3.2x.1^2+1^3\\ =\left(2x+1\right)^3\)
![](https://rs.olm.vn/images/avt/0.png?1311)
17)
\(x^3-2x^2+x\\ =x\left(x^2-2x+1\right)\\ =x\left(x-1\right)^2\)
18)
\(3\left(x+4\right)-x^2-4x\\ =3\left(x+4\right)-x\left(x+4\right)\\ =\left(x+4\right)\left(3-x\right)\)
19)
\(x^2+5x-6\\ =x^2+6x-x-6\\ =x\left(x+6\right)-\left(x+6\right)\\ =\left(x+6\right)\left(x-1\right)\)
20)
\(x^2+x-20\\ =x^2+5x-4x-20\\ =x\left(x+5\right)-4\left(x+5\right)\\ =\left(x+5\right)\left(x-4\right)\)
\(17,x^3-2x^2+x=x\left(x^2-2x+1\right)=x\left(x-1\right)^2\)
\(18,3\left(x+4\right)-x^2-4x=3\left(x+4\right)-x\left(x+4\right)=\left(x+4\right)\left(3-x\right)\)
\(19,x^2+5x-6=x^2-x+6x-6=x\left(x-1\right)+6\left(x-1\right)=\left(x-1\right)\left(x+6\right)\)
\(20,x^2+x-20=x^2-4x+5x-20=x\left(x-4\right)+5\left(x-4\right)=\left(x-4\right)\left(x+5\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a ) \(x^2+5x+6\)
\(=x^2+5x+\frac{25}{4}-\frac{1}{4}\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}\)
b ) \(x^2\left(1-x^2\right)-4+4x^2\)
\(=x^2\left(1-x^2\right)-4\left(1-x^2\right)\)
\(=\left(x^2-4\right)\left(1-x^2\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(1-x\right)\left(1+x\right)\)
a) \(x^2+5x+6\\ =x^2+5x+\frac{25}{4}-\frac{1}{4}\\ =\left(x+\frac{5}{2}\right)^2-\frac{1}{4}\\ \)
b) \(x^2\left(1-x^2\right)-4+4x^2\\ =x^2\left(1-x^2\right)-4\left(1-x^2\right)\\ =\left(x^2-4\right)\left(1-x^2\right)\\ =\left(x-2\right)\left(x+2\right)\left(1-x\right)\left(1+x\right)\)
a/ \(x^2+5x+6\)
\(=x^2+5x+\frac{25}{4}-\frac{1}{4}\)
\(=\left(x+\frac{5}{2}\right)^2-\frac{1}{4}\)
\(=\left(x+3\right)\left(x+2\right)\)
b/ \(x^2\left(1-x^2\right)-4+4x^2\)
\(=x^2\left(1-x^2\right)-4\left(1-x^2\right)\)
\(=\left(x^2-4\right)\left(1-x^2\right)\)
\(=\left(x-2\right)\left(x+2\right)\left(1-x\right)\left(1-x\right)\)
a, x2+5x=6
= x(x+5)+6
b, x2(1-x2)-4+4x2
= x2.1-x2-4+4x2
= x2(1-1-4+4)
= x2.0