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a) Ta có: \(x^2-3x+xy-3y\)
\(=x\left(x-3\right)+y\left(x-3\right)\)
\(=\left(x-3\right)\left(x+y\right)\)
b) Ta có: \(x^3+10x^2+25x-xy^2\)
\(=x\left(x^2+10x+25-y^2\right)\)
\(=x\left(x+5-y\right)\left(x+5+y\right)\)
c) Ta có: \(x^3+2+3\left(x^3-2\right)\)
\(=4x^3-4\)
\(=4\left(x-1\right)\left(x^2+x+1\right)\)
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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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\(a,=x\left(x-8\right)\\ b,=x\left(x-y\right)-6\left(x-y\right)=\left(x-6\right)\left(x-y\right)\\ c,=\left(x-3\right)^2-y^2=\left(x-y-3\right)\left(x+y-3\right)\\ d,=\left(x+y\right)\left(x^2-xy+y^2\right)+2\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2+2\right)\)
a: \(x^2-8x=x\left(x-8\right)\)
c: \(x^2-6x+9-y^2=\left(x-3-y\right)\left(x-3+y\right)\)
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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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a) \(2x-72x^3=2x\left(1-36x^2\right)=2x\left(1-6x\right)\left(1+6x\right)\)
f) \(4x^4+1=4x^4+4x^2+1-4x^2=\left(2x^2+1\right)^2-\left(2x\right)^2=\left(2x^2-2x+1\right)\left(2x^2+2x+1\right)\)
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a) x2y+xy+x+1= (x2y+xy)+(x+1)=xy(x+10+(x+1)=(x+1)(xy+1)
b) x2-(a+b)x+ab=x2-ax-bx+ab=(x2-ax)-(bx-ab)=x(x-a)-b(x-a)=(x-a)(x-b)
c) ax2+ay-bx2-by=(ax2+ay)-(bx2+by)=a(x2+y)-b(x2+y)=(a-b)(x2+y)
d) ax-2x-a2+2a=(ax-2x)-(a2-2a)=x(a-2)-a(a-2)=(a-2)(x-a)
e) 2x2+4ax+x+2a=(2x2+4ax)+(x+2a)=2x(x+2a)+(x+2a)=(x+2a)(2x+1)
f) x3+ax2+x+a=(x3+ax2)+(x+a)=x2(x+a)+(x+a)=(x2+1)(x+a)
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a: \(16x^5-25x^3\)
\(=x^3\left(16x^2-25\right)\)
\(=x^3\left(4x-5\right)\left(4x+5\right)\)
b: \(x^2-2xy+y^2-16\)
\(=\left(x-y\right)^2-16\)
\(=\left(x-y-4\right)\left(x-y+4\right)\)
c: \(x^2-5y-xy+5x\)
\(=x\left(x-y\right)+5\left(x-y\right)\)
\(=\left(x-y\right)\left(x+5\right)\)
d: \(x^2\left(x^2+4\right)-x^2+4\)
\(=\left(x^2+4\right)\left(x^2-1\right)\)
\(=\left(x^2+4\right)\left(x-1\right)\left(x+1\right)\)
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a, \(=\left(xy+1+x-y\right)\left(xy+1-x+y\right)\)
b, \(\left(x+y-x+y\right)[\left(x+y\right)^2+\left(x+y\right)\left(x-y\right)+\left(x-y\right)^2]\)
\(=2y[x^2+2xy+y^2+x^2-y^2+x^2-2xy+y^2]\)
\(=2y\left(3x^2+y^2\right)\)
c,\(=3\left(x+1\right)^2\left(x^2-x+1\right)y^2\)
câu a, b áp dụng hằng đẳng thức rồi làm nha
c) 3x4y2 + 3x3y2 + 3xy2 + 3y2
= ( 3x4y2 + 3x3y2 ) + ( 3xy2 + 3y2 )
= 3x3y2 ( x + 1) + 3y2 ( x + 1 )
= ( 3x3y2 + 3y2 ) ( x + 1 )
= 3y2 ( x3 + 1 ) ( x + 1 )
= 3y2 ( x + 1 ) ( x2 - x + 1 ) ( x + 1 )
= 3y2 ( x + 1 )2 ( x2 - x + 1 )
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c. x2 - 5x + 4
= x2 + x + 4x + 4
= x(x + 1) + 4(x + 1)
= (x + 4)(x + 1)
a) x3 - x2 - x - 2 = x3 - 2x2 + x2 - 2x + x - 2
= x2 (x-2) + x (x-2) + (x-2)
= (x-2)(x2+x+1)
\(x^3-x^2-x-2\)
\(=x^2.\left(x-1\right)-\left(x-1\right)-1^2\)
\(=\left(x-1\right)\left(x^2-1\right)-1^2\)
\(=\left(x-1\right)^2\left(x+1\right)-1^2\)
\(=\left[\left(x-1\right).\sqrt{x+1}\right]^2-1^2\)
\(=\left[\left(x-1\right).\sqrt{x+1}-1\right].\left[\left(x-1\right).\sqrt{x+1}+1\right]\)
Tham khảo nhé~