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\(x^2-5y+y^2-2xy+5x=\left(x^2-2xy+y^2\right)+\left(5x-5y\right)\)
a/ x2 – 5y + y2 -2xy + 5x = ( x2 - 2xy + y2 ) - 5( y - x ) = ( x - y )2 - 5( y - x ) = ( y - x )2 - 5( y - x ) = ( y - x )( y - x - 5 )
b/ 4x2 – 81(y – 2)2 = 4x2 - 92(y – 2)2= 4x2 – ( 9y – 18)2 = ( 2x -9y -18 )( 2x + 9y + 18 )
c/ x2z – y2z + 2yz – z = ( x2z + yz ) - ( y2z - yz ) - z = z( x2 + y ) - z( y2 - y ) -z = z( x2 + y - y2 +y - 1 ) = z( x2 + 2y - y2 - 1 ) \(=z[x^2-\left(y^2-2y+1\right)]=z[x^2-\left(y-1\right)^2=z\left(x-y+1\right)\left(x+y-1\right)\)
d/ x3 – 8y3 + x2 + 2xy + 4y2 = ( x3 – 8y3 ) + x2 + 2xy + 4y2 = ( x -2y )( x2 + 2xy + 4y2 ) + ( x2 + 2xy + 4y2 0 = ( x2 + 2xy + 4y2)( x -2y +1)
e/ 7x2 – 11x + 4 = 7x2 -7x -4x +4 = 7x( x-1 ) - 4( x - 1 ) = ( x - 1 )( 7x - 4 )
g/ 13x2 + 2xy – 15y2 = 13x2 - 13xy + 15xy - 15y2 = 13x( x - y ) + 15y( x - y ) = ( x - y )( 13x + 15y )
h/ x3 + 3x2 + 3x + 2 = x3 +2x2 + x2 +2x + x + 2 = x2( x + 2 ) + x( x + 2 ) + ( x + 2 ) = ( x + 2 )( x2 + x + 1 )
i/ x3 – 3x2 + 3x – 2 + xy – 2y = x3 - 2x2 - x2 + 2x + x - 2 +xy - 2y = x2( x - 2 ) - x( x - 2 ) + ( x - 2 ) + y( x - 2 ) = ( x - 2 )( x2 - x +1 + y )
\(\left(x^2-4\right)+\left(8-5.x\right).\left(x+2\right)+4.\left(x-2\right).\left(x+1\right)=0\)
\(\Leftrightarrow x^2-4+8.x+16-5.x^2-10.x+\left(4.x-8\right).\left(x+1\right)=0\)
\(\Leftrightarrow x^2-4+8.x+16-5.x^2-10.x+4.x^2+4.x-8.x-8=0\)
\(\Leftrightarrow0+4-6.x=0\)
\(\Leftrightarrow4-6.x=0\)
\(\Leftrightarrow-6.x=-4\)
\(\Rightarrow x=\frac{2}{3}\)
Vậy x = \(\frac{2}{3}\)
Trả lời:
a, \(-xy.\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+3xy\)
b, \(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y\)
\(=12x^6y^5:6x^2y^2-3x^3y^4:6x^2y+4x^2y+6x^2y\)
\(=2x^4y^3-\frac{1}{2}xy^3+\frac{2}{3}\)
a.\(\left(-xy\right)\left(x^2+2xy-3\right)=-x^3y-2x^2y^2+6xy\)
b.\(\left(12x^6y^5-3x^3y^4+4x^2y\right):6x^2y=2x^4y^4-\frac{1}{2}xy^3+\frac{2}{3}\)