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a) 6x² + 7xy + 2y²
= 6x² + 4xy + 3xy + 2y²
= (6x² + 4xy) + (3xy + 2y²)
= 2x(3x + 2y) + y(3x + 2y)
= (3x + 2y)(2x + y)
b) x² - y² + 10x - 6y + 16
= x² + 10x + 25 - y² - 6y - 9
= (x² + 10x + 25) - (y² + 6y + 9)
= (x + 5)² - (y + 3)²
= (x + 5 - y - 3)(x + 5 + y + 3)
= (x - y + 2)(x + y + 8)
c) 4x⁴ + y⁴
= 4x⁴ + 4x²y² + y⁴ - 4x²y²
= (2x² + y²)² - (2xy)²
= (2x² + y² - 2xy)(2x² + y² + 2xy)
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\(3x^2-7xy+2y^2\)
\(=3x^2-6xy-xy+2y^2\)
\(=\left(3x^2-6xy\right)-\left(xy-2y^2\right)\)
\(=3x\left(x-2y\right)-y\left(x-2y\right)\)
\(=\left(x-2y\right)\left(3x-y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=6x^2-9xy+2xy-3y^2\)
=3x(2x-y)+y(2x-y)
=(2x-y)(3x+y)
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b: \(=\left(x-5\right)^2-9y^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
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Bài 1:
b: \(=\left(x-5\right)^2-9y^2\)
\(=\left(x-5-3y\right)\left(x-5+3y\right)\)
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\(1,\\ a,=3x\left(x-3y\right)\\ b,=\left(x-5\right)^2-9y^2=\left(x-3y-5\right)\left(x+3y-5\right)\\ c,=3x\left(x-y\right)-2\left(x-y\right)=\left(3x-2\right)\left(x-y\right)\\ 2,\\ Sửa:x^2-6x+10=\left(x-3\right)^2+1\ge1>0,\forall x\)
1, =3x (2x -3y)
c, = 3x(x-y) -2(x-y)
= (3x-2)(x-y)
2, Ta có: x2 -6x+10= (x-3)2 +11
Nhận xét: (x-3)2 >= 0 với mọi số thực x
=> (x-3)2 +1 >= 1 >0 (đpcm)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(=-2\left(x^2-2xy+y^2-4\right)\)
\(=-2\left[\left(x-y\right)^2-4\right]\)
\(=-2\left(x-y-2\right)\left(x-y+2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
\(6x^2-7x+2\)
\(=6x^2-3x-4x+2\)
\(=\left(6x^2-3x\right)-\left(4x-2\right)\)
\(=3x\left(2x-1\right)-2\left(2x-1\right)\)
\(=\left(2x-1\right)\left(3x-2\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
2x2 + 4x + 2 – 2y2 (có nhân tử chung là 2)
= 2.(x2 + 2x + 1 – y2) (Xuất hiện x2 + 2x + 1 là hằng đẳng thức)
= 2[(x2 + 2x + 1) – y2]
= 2[(x + 1)2 – y2] (Xuất hiện hằng đẳng thức (3))
= 2(x + 1 – y)(x + 1 + y)
6x2+7xy+2y2
=6x2+3xy+4xy+2y2
=3x.(2x+y)+2y.(2x+y)
=(2x+y)(3x+2y)