Câu hỏi của Đặng Thị Ngọc Anh

Question

Phân tích đa thức sau thành nhân tử : (x+y+z)^3 -x^3-y^3-z^3(Ko làm tắt giùm ạ)

l҉o҉n҉g҉ d҉z҉ · Accepted Answer

( x + y + z )3 - x3 - y3 - z3= [ ( x + y + z )3 - x3 ] - ( y3 + z3 )= ( x + y + z - x )[ ( x + y + z )2 + ( x + y + z )x + x2 ] - ( y + z )( y2 - yz + z2 )= ( y + z )( 3x2 + y2 + z2 + 2yz + 3zx + 3xy ) - ( y + z )( y2 - yz + z2 )= ( y + z )( 3x2 + y2 + z2 + 2yz + 3zx + 3xy - y2 + yz - z2 )= ( y + z )( 3x2 + 3yz + 3zx + 3xy )= 3( y + z )( x2 + yz + zx + xy )= 3( y + z )[ ( x2 + zx ) + ( xy + yz ) ]= 3( y + z )[ x( x + z ) + y( x + z ) ]= 3( y + z )( x + z )( x + y )Câu trả lời: ( x + y + z )3 - x3 - y3 - z3= [ ( x + y + z )3 - x3 ] - ( y3 + z3 )= ( x + y + z - x )[ ( x + y + z )2 + ( x + y + z )x + x2 ] - ( y + z )( y2 - yz + z2 )= ( y + z )( 3x2 + y2 + z2 + 2yz + 3zx + 3xy ) - ( y + z )( y2 - yz + z2 )= ( y + z )( 3x2 + y2 + z2 + 2yz + 3zx + 3xy - y2 + yz - z2 )= ( y + z )( 3x2 + 3yz + 3zx + 3xy )= 3( y + z )( x2 + yz + zx + xy )= 3( y + z )[ ( x2 + zx ) + ( xy + yz ) ]= 3( y + z )[ x( x + z ) + y( x + z ) ]= 3( y + z )( x + z )( x + y )

Greninja · Answer

$\left(x+y+z\right)^3-x^3-y^3-z^3$$=\left[\left(x+y+z\right)^3-x^3\right]-\left(y^3+z^3\right)$$=\left(x+y+z-x\right).\left[\left(x+y+z\right)^2+\left(x+y+z\right).x+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)$$=\left(y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2\right)-\left(y+z\right)\left(y^2-yz+z^2\right)$$=\left(y+z\right).\left[x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2-\left(y^2-yz+z^2\right)\right]$$=\left(y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2-y^2+yz-z^2\right)$$=\left(y+z\right).\left(3x^2+3xy+3yz+3xz\right)$$=\left(y+z\right).\left[\left(3x^2+3xy\right)+\left(3yz+3xz\right)\right]$$=\left(y+z\right).\left[3x.\left(x+y\right)+3z.\left(y+x\right)\right]$$=\left(y+z\right).\left(x+y\right).\left(3x+3z\right)$$=3.\left(y+z\right).\left(x+y\right).\left(x+z\right)$Câu trả lời: $\left(x+y+z\right)^3-x^3-y^3-z^3$$=\left[\left(x+y+z\right)^3-x^3\right]-\left(y^3+z^3\right)$$=\left(x+y+z-x\right).\left[\left(x+y+z\right)^2+\left(x+y+z\right).x+x^2\right]-\left(y+z\right)\left(y^2-yz+z^2\right)$$=\left(y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2\right)-\left(y+z\right)\left(y^2-yz+z^2\right)$$=\left(y+z\right).\left[x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2-\left(y^2-yz+z^2\right)\right]$$=\left(y+z\right).\left(x^2+y^2+z^2+2xy+2yz+2xz+x^2+yx+zx+x^2-y^2+yz-z^2\right)$$=\left(y+z\right).\left(3x^2+3xy+3yz+3xz\right)$$=\left(y+z\right).\left[\left(3x^2+3xy\right)+\left(3yz+3xz\right)\right]$$=\left(y+z\right).\left[3x.\left(x+y\right)+3z.\left(y+x\right)\right]$$=\left(y+z\right).\left(x+y\right).\left(3x+3z\right)$$=3.\left(y+z\right).\left(x+y\right).\left(x+z\right)$

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