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a) = (4x2 +4x + 1) - (9y2 + 6y + 1) = (2x+ 1)2 - (3y + 1)2 = (2x+ 1+ 3y + 1).(2x + 1 - 3y - 1) = (2x+ 3y+ 2).(2x-3y)
b) = x2 - 2015x + 2014x - 2014.2015 = x.(x - 2015) + 2014. (x - 2015) = (x + 2014).(x - 2015)
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\(25-4x^2+12xy-9y^2\)
\(=25-\left(2x-3y\right)^2\)
\(=\left(5-2x+3y\right)\left(5+2x-3y\right)\)
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x2 + 4x + y - 9y2
<=> x(x + 4) + y(1 + 9y)
<=> (x + y)(x + 4 + 1 + 9y)
<=> (x + y)(x + 9y + 5)
bí rồi
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\(x^2-y^2+4-4x\)
\(=\left(x^2-4x+4\right)-y^2\)
\(=\left(x-2\right)^2-y^2\)
\(=\left(x-2+y\right)\left(x-2-y\right)\)
1)=x(x-1)-y(y-1)
2)=(x-2)2 -y2
3)=(2x+1)2 -9y2+1
#Mình k biết viết bình phương, thông cảm bạn nhé!
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\(a.=x^3+3x^2y+3x^2y+9xy^2+3xy^2+9y^3\)
\(=x^2\left(x+3y\right)+3xy\left(x+3y\right)+3y^2\left(x+3y\right)\)
\(=\left(x+3y\right)\left(x^2+3xy+3y^2\right).\)
\(b.=9x^3+3x^2y+9x^2y+3xy^2+3xy^2+y^3\)
\(=3x^2\left(3x+y\right)+3xy\left(3x+y\right)+y^2\left(3x+y\right)\)
\(=\left(3x^2+3xy+y^2\right)\left(3x+y\right)\).
1) Phân tích đa thức thành nhân tử
a) \(x^2+2x+1-y^2\)
b) \(x^2+4x+3\)
c) \(4x^2-9y^2\)
d) \(x^3-27y^3\)
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a) \(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
b) \(x^2+4x+3=x^2+4x+4-1=\left(x+2\right)^2-1=\left(x+1\right)\left(x+3\right)\)
c) \(4x^2-9y^2=\left(4x-9y\right)\left(4x+9y\right)\)
d) \(x^3-27y^3=\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
a)\(x^2+2x+1-y^2=\left(x+1\right)^2-y^2=\left(x+y-1\right)\left(x-y+1\right)\)
b)\(x^2+4x+3=\left(x+1\right)\left(x+3\right)\)
c)\(4x^2-9y^2=\left(2x-3y\right)\left(2x+3y\right)\)
d)\(x^3-27y^3=\left(x-3y\right)\left(x^2+3xy+9y^2\right)\)
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a,x2-4xy+4y2
=(x-2y2
b,4x4+9y2-12x2y
=(2x2)2+(3y)2-12x2y
(2x2-3y)
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\(e,-5x+x^2-14\)
\(=x^2+2x-7x-14\)
\(=x\left(x+2\right)-7\left(x+2\right)\)
\(=\left(x+2\right)\left(x-7\right)\)
\(f,x^3+8+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+2x+4\right)+6x\left(x+2\right)\)
\(=\left(x+2\right)\left(x^2+8x+4\right)\)
\(g,15x^2-7xy-2y^2\)
\(=15x^2+3xy-10xy-2y^2\)
\(=3\left(5x+y\right)-2y\left(5x+y\right)\)
\(=\left(5x+y\right)\left(3-2y\right)\)
\(h,3x^2-16x+5\)
\(=3x^2-x-15x+5\)
\(=x\left(3x-1\right)+5\left(3x-1\right)\)
\(=\left(3x-1\right)\left(x+5\right)\)
\(a,x^3+2x^2y+xy^2=x\left(x^2+2xy+y^2\right)\)
\(=x\left(x+y\right)^2\)
\(b,4x^2-9y^2+4x-6y\)
\(=4x^2+4x+1-\left(9y^2+6y+1\right)\)
\(=\left(2x+1\right)^2-\left(3y+1\right)^2\)
\(=\left(2x-3y\right)\left(2x+3y+2\right)\)
\(c,-x^2+5x+2xy-5y-y^2\)
\(=-\left(x^2-2xy+y^2\right)+5\left(x-y\right)\)
\(=-\left(x-y\right)^2+5\left(x-y\right)\)
\(=\left(x-y\right)\left(y-x+5\right)\)
\(d,x^2+4x-12\)
\(=x^2-2x+6x-12\)
\(=x\left(x-2\right)+6\left(x-2\right)\)
\(=\left(x-2\right)\left(x+6\right)\)
hello Đam Mê Ngành Luật:) sửa +3 thành -3
\(4x^2+9y^2-12xy+4x-6y-3\)
\(=\left(4x^2-12xy+9y^2+4x-6y+1\right)-4\)
\(=\left(2x-3y+1\right)^2-2^2=\left(2x-3y-1\right)\left(2x-3y+3\right)\)
Đề vẫn đúng là +3 bae à :<< cậu xem lại xem