phân tích đa thức thành phân tử 3xy2-12xy+12x ?
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\(12x-9-4x^2=-\left(2x-3\right)^2\\ Sửa:x^3-6x^2y+12xy^2-8y^3=\left(x-2y\right)^3\)
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(12x^2 - 12xy + 3y^2) - 10.(2x - y) + 8
= 3(4x^2 - 4xy + y^2) - 10(2x - y) + 8
= 3(2x - y)^2 - 10(2x - y) + 8
= 3(2x - y)^2 - 10(2x - y) + 8
= 3(2x - y)^2 - 6(2x - y) - 4(2x - y) + 8
= 3(2x - y)(2x - y - 2) - 4(2x - y -2)
= (2x - y -2)[3(2x - y) - 4]
= (2x - y -2)(6x - 3y -4)
Ai k mk mk k lại
(12x^2 - 12xy + 3y^2) - 10.(2x - y) + 8
= 3(4x^2 - 4xy + y^2) - 10(2x - y) + 8
= 3(2x - y)^2 - 10(2x - y) + 8
= 3(2x - y)^2 - 10(2x - y) + 8
= 3(2x - y)^2 - 6(2x - y) - 4(2x - y) + 8
= 3(2x - y)(2x - y - 2) - 4(2x - y -2)
= (2x - y -2)[3(2x - y) - 4]
= (2x - y -2)(6x - 3y -4)
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3*(\(4x^2-4xy+y^2\))-10(2x-y)+8
3*(2x-y)^2-10(2x-y)+8
3*(2x-y)^2-6(2x-y)-4(2x-y)+8
3(2x-y)(2x-y-2)-4(2x-y-2)
(2x-y-2)(6x-3y-40
\(\left(12x^2-12xy+3y^2\right)-10\left(2x-y\right)+8\)
\(=\left(12x^2-6xy-6xy+3y^2\right)-10\left(2x-y\right)+8\)
\(=\left[6x\left(2x-y\right)-3y\left(2x-y\right)\right]-10\left(2x-y\right)+8\)
\(=\left(2x-y\right)\left(6x-3y\right)-10\left(2x-y\right)+8\)
\(=3\left(2x-y\right)^2-10\left(2x-y\right)+8\)
Đặt \(2x-y=a\), khi đó biểu thức có dạng:
\(3a^2-10a+8=3a^2-6a-4a+8\)
\(=3a\left(a-2\right)-4\left(a-2\right)=\left(a-2\right)\left(3a-4\right)\)
\(=\left(2x-y-2\right)\left(6x-3y-4\right).\)
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a. 3xy( 4x + y - \(\dfrac{4}{3}\) )
b. 2x2( 3x + 1 )
c. (2x + 3 )( x - y )
d. xy( 1 - x )( x - 1 )
e. 6( 2x + 1 )( x + y )
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\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
\(x^3-x+3x^2+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
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3xy^2-12xy+12x
=3x(y^2-4y+4)
=3x(y-2)^2