Phân tích đa thức thành nhân tử :
\(\left(12x^2-12xy+3y^2\right)-10\left(2x-y\right)+8\)
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a) \(A=x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2-1+3x-3y-3\)
\(=\left(x-y-1\right)\left(x-y+1\right)+3\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x-y+1+3\right)\)
\(=\left(x-y-1\right)\left(x-y+4\right)\)
(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)
\(\left(2x-y\right)\left(x-y\right)-\left(3y-4x\right)^2+\left(y-2x\right)\left(2y-3x\right)\)
=(2x-y)(x-y)-(2x-y)(2y-3x)-(4x-3y)2
=(2x-3y)(x-y-2y+3x)-(4x-3y)2
=(2x-3y)(4x-3y)-(4x-3y)2
=(4x-3y)(2x-3y-4x+3y)
=(4x-3y))(-2x)
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:
\(3a^2-10a+8=3a^2-6a-4a+8\)
\(=3a\left(a-2\right)-4\left(a-2\right)=\left(a-2\right)\left(3a-4\right)\)
\(=\left(2x-y-2\right)\left(6x-3y-4\right).\)