Phân tích đa thức thành nhân tử: x6-y3
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a) (x - y)(x + y + 3). b) (x + y - 2xy)(2 + y + 2xy).
c) x 2 (x + l)( x 3 - x 2 + 2). d) (x – 1 - y)[ ( x - 1 ) 2 + ( x - 1 ) y + y 2 ].
a) x⁶ + y⁶ = (x²)³ + (y²)³
= (x² + y²)(x⁴ - x²y² + y⁴)
b) x⁶ - y⁶
= (x³)² - (y³)²
= (x³ - y³)(x³ + y³)
= (x - y)(x² + xy + y²)(x + y)(x² - xy + y²)
\(x^3+y^3+3y^2+3y+1\\ =x^3+\left(y+1\right)^3\\ =\left(x+y+1\right)\left[x^2-x\left(y+1\right)+\left(y+1\right)^2\right]\\ =\left(x+y+1\right)\left(x^2-xy-x+y^2+2y+1\right)\\ =\left(x+y+1\right)\left(x^2+y^2+2y+1-xy-x\right)\)
\(Sửa:x^3+y^3+2x^2+2xy\\ =\left(x+y\right)\left(x^2-xy+y^2\right)+2x\left(x+y\right)\\ =\left(x+y\right)\left(x^2-xy+y^2+2x\right)\)
x + y + z 3 - z 3 - y 3 - z 3 = ( x + y ) + z 3 – x 3 – y 3 – z 3 = ( x + y ) 3 + 3 ( x + y ) 2 z + 3 ( x + y ) z 2 + z 3 – x 3 – y 3 – z 3 = x 3 + y 3 + 3 x y ( x + y ) + 3 ( x + y ) 2 z + 3 ( x + y ) z 2 – x 3 – y 3 ( v ì z 3 – z 3 = 0 ; 3 x 2 y + 3 x y 2 = 3 x y ( x + y ) ) = 3 x y . ( x + y ) + 3 ( x + y ) 2 . z + 3 ( x + y ) . z 2 = 3 ( x + y ) [ x y + ( x + y ) z + z 2 ] = 3 ( x + y ) [ x y + x z + y z + z 2 ] = 3 ( x + y ) [ x ( y + z ) + z ( y + z ) ] = 3 ( x + y ) ( y + z ) ( x + z )
Ta có
8 x 3 + 12 x 2 y + 6 x y 2 + y 3 = ( 2 x ) 3 + 3 . ( 2 x ) 2 y + 3 . 2 x . y 2 + y 3 = ( 2 x + y ) 3
Đáp án cần chọn là: B
\(=\left(x+y\right)^3-\left(x+y\right)=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
\(x^3-x+3x^2+3xy^2+y^3-y\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x+y+1\right)\left(x+y-1\right)\)
d) (y3 + 8) + ( y2 – 4) =(y3 + 23) + ( y2 – 22)
= (y + 2)(y2 – 2y + 4) + (y + 2)( y – 2)
= (y + 2)(y2 – 2y + 4 + y – 2) = (y + 2)(y2 – y + 2)
b) x2y + 4xy + 4y – y3
= y(x2 + 4x + 4 - y2)
= y[(x2 + 4x + 4) - y2]
= y[(x + 2)2 - y2]
= y(x + 2 + y)(x + 2 - y)
\(x^6-y^3=\left(x^2\right)^3-y^3=\left(x^2-y\right)\left(x^4+x^2y+y^2\right)\)