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Bạn hội con bò gì đó ơi cho mk tham gia đc không vì là hội học hành nên .....
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a, x4 - 5x2 + 4
= x4 - 4x2 - x2 + 4
= x2 . (x2 - 4) - (x2 - 4)
= (x2 - 4) . (x2 - 1)
= (x - 2) . (x + 2) . (x - 1) . (x + 1)
a)x4-5x2+4=x4-x2-4x2+4
=(x4-x2)-(4x2-4)
=x2(x2-1)-4(x2-1)
=(x2-1)(x2-4)
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\(x^4y^4+4x^2y^2+4-4x^2y^2=\left(x^2y^2+2\right)^2-\left(2xy\right)^2=\left(x^2y^2-2xy+2\right)\left(x^2y^2+2xy+2\right)\)
\(4x^4y^4+4x^2y^2+1-4x^2y^2=\left(2x^2y^2+1\right)^2-\left(2xy\right)^2=\left(2x^2y^2-2xy+1\right)\left(2x^2y^2+2xy+1\right)\)
1. \(x^4y^4\)+4= \(\left(x^2y^2\right)^2\)+4\(x^2y^2\)+4-4\(x^2y^2\)=(\(x^2y^2\)+2)^2 - 4x^2y^2=(x^2y^2+2-2xy)(X^2y^2+2+2xy) .Câu b tương tự bạn tự làm nhé!
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a)\(9x^2y^3-3x^4y^2-6x^3y^2+18xy^4=3xy^2\left(3xy-x^3-2x^2+6y^2\right)\)
b)\(a^3x^2y^2-\frac{5}{2}a^3x^4+\frac{3}{2}a^4x^2y=a^3x^2\left(y^2-\frac{5}{2}x^2+\frac{3}{2}ay\right)\)
c)\(x^2+4xy-21y^2=\left(x^2+4xy+4y^2\right)-25y^2=\left(x+2y\right)^2-\left(5y\right)^2=\left(x+2y-5y\right)\left(x+2y+5y\right)=\left(x-3y\right)\left(x+7y\right)\)d)\(2x^4+4=2\left(x^4+4\right)=2\left(x^4+4x^2+4-4x^2\right)=2\left[\left(x^2+2\right)^2-\left(2x\right)^2\right]=2\left(x^2-2x+2\right)\left(x^2+2x+2\right)\)
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bạn tham khảo 1 số bài rồi tự làm nhé
a) 3x−3+5(1−x)
=3x−3+5−5x
=3x−5x+2
=x(3−5)+2
=−2x+2
=2(1−x)
b) 12a2−3ab+8ac−2bc
=3a(4a−b)+2c(4a−b)
=(4a−b)(3a+2c)
c) x2−25+y2−2xy
=x2−2xy+y2−25
=(x−y)2−52
=(x−y−5)(x−y+5)
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a) x4 - x3y - x + y
= x3(x - y) - (x - y)
=(x - y)(x3 - 1)
=(x - y)(x - 1)(x2 + x + 1)
b) x3 - 4x2 - 8x + 8
= (x3 + 8) - (4x2 + 8x)
= (x + 2)(x2 - 2x + 4) - 4x(x + 2)
= (x + 2)(x2 - 2x + 4 - 4x)
= (x + 2)(x2 - 6x + 4)
a, = (x^4-x^3y)-(x-y)
=x^3.(x-y)-(x-y) = (x-y).(x^3-1) = (x-y).(x-1).(x^2+x+1)
b, = (x^3+2x^2)-(6x^2+12x)+(4x+8)
= x^2.(x+2)-6x.(x+2)+4.(x+2) = (x+2).(x^2-6x+4)
k mk nha
a) \(\left(a+b\right)^3-a^3-b^3\)
\(=a^3+3a^2b+3ab^2+b^3-a^3-b^3\)
\(=3a^2b+3ab^2\)
\(=3ab\left(a+b\right)\)
b) \(\left(x+y\right)^4+x^4+y^4\)
\(=x^4+4x^3y+6x^2y^2+4xy^3+y^4+x^4+y^4\)
\(=2x^4+2y^4+4x^2y^2+4x^3y+4xy^3+2x^2y^2\)
\(=2\left(x^4+2x^2y^2+y^4\right)+4xy\left(x^2+y^2\right)+2x^2y^2\)
\(=2\left[\left(x^2+y^2\right)^2+2xy\left(x^2+y^2\right)+x^2y^2\right]\)
\(=2\left(x^2+y^2+xy\right)^2\)