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a: \(=4xy\left(1-5x^2y\right)\)
b: \(=\left(x-y\right)\left(x+y\right)+3\left(x-y\right)=\left(x-y\right)\left(x+y+3\right)\)
c: \(=x\left(x-a\right)+y\left(x-a\right)=\left(x-a\right)\left(x+y\right)\)
d: \(=\left(x+2y\right)^2-36=\left(x+2y+6\right)\left(x+2y-6\right)\)
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\(a.x^3-2x^2-2x-4\\ =\left(x^3-2x^2\right)-\left(2x-4\right)\\ =x^2\left(x-2\right)-2\left(x-2\right)\\ =\left(x^2-2\right)\left(x-2\right)\)
\(b.xy+1-x-y\\ =\left(xy-x\right)+\left(-y+1\right)\\ =x\left(y-1\right)-\left(y-1\right)\\ =\left(x-1\right)\left(y-1\right)\)
\(c.x^2-4xy+4y^2-4y\\ =\left(x-2y\right)^2-4y\\ =\left(x-2y\right)^2-\left(2y\right)^2\\ =\left(x-2y+2y\right)\left(x-2y-2y\right)\\ =x\left(x-4y\right)\)
\(d.16-x^2+2xy-y^2\\ =4^2-\left(x-y\right)^2\\ =\left(4-x+y\right)\left(4-x-y\right)\)
b: =xy-x-y+1
=x(y-1)-(y-1)
=(x-1)(y-1)
c: =(x-2y)^2-4y
\(=\left(x-2y-2\sqrt{y}\right)\left(x-2y+2\sqrt{y}\right)\)
d: =16-(x^2-2xy+y^2)
=16-(x-y)^2
=(4-x+y)(4+x-y)
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1, \(a^6+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)
2, \(x^2-10x+25=\left(x-5\right)^2\)
3, \(8x^3-\dfrac{1}{8}=\left(2x-\dfrac{1}{2}\right)\left(4x^2+x+\dfrac{1}{4}\right)\)
4, \(x^2+4xy+4y^2=\left(x+2y\right)^2\)
1) \(a^6+b^3=\left(a^2\right)^3+b^3=\left(a^2+b\right)\left(a^4-a^2b+b^2\right)\)
2) \(x^2-10x+25=\left(x-5\right)^2\)
3) \(8x^3-\dfrac{1}{8}=\left(2x\right)^3-\left(\dfrac{1}{3}\right)^3=\left(2x-\dfrac{1}{3}\right)\left(4x^2+\dfrac{2x}{3}+\dfrac{1}{4}\right)\)
4) \(x^2+4xy+4y^2=\left(x+2y\right)^2\)
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a: \(=x\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-1\right)\)
b: \(=25-\left(x-2y\right)^2\)
\(=\left(5-x+2y\right)\left(5+x-2y\right)\)
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Câu 1:
$x^2+4y^2+4xy-16=[x^2+(2y)^2+2.x.2y]-16$
$=(x+2y)^2-4^2=(x+2y-4)(x+2y+4)$
Câu 2:
$x^3+x^2+y^3+xy=(x^3+y^3)+(x^2+xy)$
$=(x+y)(x^2-xy+y^2)+x(x+y)=(x+y)(x^2-xy+y^2+x)$
Câu 1:
\(x^2+4y^2+4xy-16\)
\(=\left(x+2y\right)^2-16\)
\(=\left(x+2y+4\right)\left(x+2y-4\right)\)
Câu 2:
\(x^3+x^2+y^3+xy\)
\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+x\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+x\right)\)
![](https://rs.olm.vn/images/avt/0.png?1311)
a) Áp dụng HĐT 1 thu được ( 2 x + y ) 2 .
b) Áp dụng HĐT 3 với A = 2x + l; B = x - l thu được
[(2x +1) + (x -1)] [(2x +1) - (x -1)] rút gọn thành 3x(x + 2).
c) Ta có: 9 - 6x + x 2 - y 2 = ( 3 - x ) 2 - y 2 = (3 - x - y)(3 -x + y).
d) Ta có: -(x + 2) + 3( x 2 - 4) = -{x + 2) + 3(x + 2)(x - 2)
= (x + 2) [-1 + 3(x - 2)] = (x + 2)(3x - 7).
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= \(-\left(x^2+4xy+4y^2\right)\)
= \(-\left(x+2y\right)^2\)
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a) \(9x^2-16\)
\(=\left(3x\right)^2-4^2\)
\(=\left(3x-4\right)\left(3x+4\right)\)
b) \(x^2+4xy+4y^2-3x-6y\)
\(=\left(x^2+4xy+4y^2\right)-\left(3x+6y\right)\)
\(=\left[x^2+2\cdot x\cdot2y+\left(2y\right)^2\right]-3\left(x+2y\right)\)
\(=\left(x+2y\right)^2-3\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x+2y-3\right)\)
#\(Toru\)
a) xy + y2 - x - y
= ( xy – x ) + ( y^2 – y )
= x (y – 1) + y (y – 1)
= (y – 1) (x + y)
b) 25 – x^2 + 4xy - 4y^2
= 5^2 – (x^2 – 4xy + 4y^2)
= 5^2 – (x – 2y)^2
= (5 – x + 2y)(5 + x – 2y)
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