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a) \(\left(5x-4\right)^2-49x^2\)
\(=\left(5x-4\right)^2-\left(7x\right)^2\)
\(=\left(12x-4\right)\left(-2x-4\right)\)
\(=-6\left(3x-1\right)\left(x+2\right)\)
c) \(x^2-y^2-x+y\)
\(=\left(x+y\right)\left(x-y\right)-\left(x-y\right)\)
\(=\left(x+y-1\right)\left(x-y\right)\)
d)\(4x^2-9y^2+4x-6y\)
\(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2y-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
e) \(-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+\left(5x-5y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(y-x+5\right)\)
f) \(y^2\left(x^2+y\right)-zx^2-zy\)
\(=y^2\left(x^2+y\right)-z\left(x^2+y\right)\)
\(=\left(y^2-z\right)\left(x^2+y\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Xem lại đề
b) x³ - 4x²y + 4xy² - 9x
= x(x² - 4xy + 4y² - 9)
= x[(x² - 4xy + 4y² - 3²]
= x[(x - 2y)² - 3²]
= x(x - 2y - 3)(x - 2y + 3)
c) x³ - y³ + x - y
= (x³ - y³) + (x - y)
= (x - y)(x² + xy + y²) + (x - y)
= (x - y)(x² + xy + y² + 1)
d) 4x² - 4xy + 2x - y + y²
= (4x² - 4xy + y²) + (2x - y)
= (2x - y)² + (2x - y)
= (2x - y)(2x - y + 1)
e) 9x² - 3x + 2y - 4y²
= (9x² - 4y²) - (3x - 2y)
= (3x - 2y)(3x + 2y) - (3x - 2y)
= (3x - 2y)(3x + 2y - 1)
f) 3x² - 6xy + 3y² - 5x + 5y
= (3x² - 6xy + 3y²) - (5x - 5y)
= 3(x² - 2xy + y²) - 5(x - y)
= 3(x - y)² - 5(x - y)
= (x - y)[(3(x - y) - 5]
= (x - y)(3x - 3y - 5)
![](https://rs.olm.vn/images/avt/0.png?1311)
a: 2x^2y-50xy=2xy(x-25)
b: 5x^2-10x=5x(x-2)
c: 5x^3-5x=5x(x^2-1)=5x(x-1)(x+1)
d: \(x^2-xy+x=x\left(x-y+1\right)\)
e: x(x-y)-2(y-x)
=x(x-y)+2(x-y)
=(x-y)(x+2)
f: 4x^2-4xy-8y^2
=4(x^2-xy-2y^2)
=4(x^2-2xy+xy-2y^2)
=4[x(x-2y)+y(x-2y)]
=4(x-2y)(x+y)
f1: x^2ỹ-y^2+y
=(x-y)(x+y)+(x+y)
=(x+y)(x-y+1)
![](https://rs.olm.vn/images/avt/0.png?1311)
5x^2+10xy+5y^2
=5.(x2+2xy+y2)
=5.(x+y)2
x^3-6x^2+9x
=x.(x2-6x+9)
=x.(x-3)2
xy+y^2-x-y
=y.(x+y)-(x+y)
=(x+y)(y-1)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) \(\left(5x+4\right)^2-49x^2\)
\(=\left(5x+4-7x\right)\left(5x+4+7x\right)\)
\(=\left(4-2x\right)\left(12x+4\right)\)
\(=8\left(2-x\right)\left(3x+1\right)\)
b) \(x^3+2x^2+xy^2\)
\(=x\left(x^2+2x+y^2\right)\)
\(=x\left(x+y\right)^2\)
c)\(x^2-y^2-x+y\)
\(=\left(x-y\right)\left(x+y\right)-\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y-1\right)\)
d) \(4x^2-9y^2+4x-6y\)
\(=\left(2x-3y\right)\left(2x+3y\right)+2\left(2x-3y\right)\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
e) \(-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+\left(5x-5y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=-\left(x-y\right)\left(x-y-5\right)\)
f) \(y^2\left(x^2+y\right)-zx^2-zy\)
\(=y^2\left(x^2+y\right)-z\left(x^2+y\right)\)
\(=\left(x^2+y\right)\left(y^2-z\right)\)
\(=\left(x^2+y\right)\left(y-\sqrt{z}\right)\left(y+\sqrt{z}\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
à, bạn ơi, bạn có hể cho mình biết số mũ ở đâu được ko, mình nhìn đề ko biết ở đâu có mũ cả ....
chủ nhật nghỉ, mới ngủ z, làm bài cho tỉnh
a) = 2xy(x2 -3y)
b) = (2x-1)2
c) = x(x-2y) - 3(x-2y) = (x-2y)(x-3)
.........mỏi tay quá
![](https://rs.olm.vn/images/avt/0.png?1311)
a, \(5y^2-5x^2+6x+6y=5\left(y-x\right)\left(x+y\right)+6\left(x+y\right)\)
\(=\left(x+y\right)\left(5y-5x+6\right)\)
b, \(12x^2+19x+7=12x^2+12x+7x+7\)
\(=12x\left(x+1\right)+7\left(x+1\right)=\left(12x+7\right)\left(x+1\right)\)