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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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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 ) \(2x-1-x^2\)
\(=\left(x-1\right)-\left(x^2-x\right)\)
\(=\left(x-1\right)\left(1-x\right)\)
\(=-\left(x-1\right)^2\)
b) \(8x^3+y^6\)
\(=\left(2x+y^2\right)\left(4x^2-2xy^2+y^4\right)\)
c) \(x^2-16+4xy+4y^2\)
\(=\left(x^2+4xy+4y^2\right)-16\)
\(=\left(x+2y\right)^2-16\)
\(=\left(x+2y+4\right)\left(x+2y-4\right)\)
a) x2 + 6x + 9 = x2 + 2 . x . 3 + 32 = (x + 3)2
b) 10x – 25 – x2 = -(-10x + 25 +x2) = -(25 – 10x + x2)
= -(52 – 2 . 5 . x – x2) = -(5 – x)2
c) 8x3 - 1/8 = (2x)3 – (1/2)3 = (2x - 1/2)[(2x)2 + 2x . 12 + (1/2)2]
= (2x - 1/2)(4x2 + x + 1/4)
d)1/25x2 – 64y2 = (1/5x)2(1/5x)2- (8y)2 = (1/5x + 8y)(1/5x - 8y)
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\(\left(x^2+4y^2-20\right)^2-16\left(xy-4\right)^2=\left(x^2+4y^2-20\right)^2-\left(4xy-16\right)^2=\left(x^2+4y^2-20-4xy+16\right)\left(x^2+4y^2-20+4xy-16\right)=\left[\left(x-2y\right)^2-4\right]\left[\left(x+2y\right)^2-36\right]=\left(x-2y-2\right)\left(x-2y+2\right)\left(x+2y-6\right)\left(x+2y+6\right)\)
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b: \(=\left(x-y\right)^2-4y^2\)
\(=\left(x-y-2y\right)\left(x-y+2y\right)\)
\(=\left(x-3y\right)\left(x+y\right)\)
c: \(=x\left(x-6\right)+y\left(x-6\right)\)
\(=\left(x-6\right)\left(x+y\right)\)
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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:
\(x^3+x^2+y^3+xy\)
\(=\left(x^3+y^3\right)\left(x^2+xy\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2\right)+x\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2-xy+y^2+x\right)\)
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= \(-\left(x^2+4xy+4y^2\right)\)
= \(-\left(x+2y\right)^2\)
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Ta thấy: \(x^2+4y^2+4xy=x^2+\left(2y\right)^2+2\cdot x\cdot2y=\left(x+2y\right)^2\)
câu trả lời hay nhất đấy
x^2 + 4xy - 16 + 4y^2
= x^2 + 4xy + 4y^2 - 4^2
= (x + 2y)^2 - 4^2
= (x + 2y - 4)(x + 2y + 4)
2x^2-5xy-3y^2
= 2^x + xy - 6xy - 3y^2
= x(2x + y) - 3y(2x + y)
= (2x + y)(x - 3y)
a,\(x^2-2x+2y-xy\)
\(=\left(x^2-xy\right)+\left(-2x+2y\right)\)
\(=x\left(x-y\right)-2\left(x-y\right)\)
\(=\left(x-2\right)\left(x-y\right)\)
a. \(x^2-2x+2y-xy=\left(x^2-2x\right)-\left(xy-2y\right)=x\left(x-2\right)-y\left(x-2\right)=\left(x-2\right)\left(x-y\right)\)
b. \(x^2+4xy-16+4y^2=\left(x^2+4xy+4y^2\right)-16=\left(x+2y\right)^2-4^2=\left(x+2y-4\right)\left(x+2y+4\right)\)
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