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\(x^2+4y^2+3x-6y=\left(x^2+3x\right)-\left(4y^2+6y\right)=x\left(x+3\right)-2y\left(2y+3\right)\)
x^3 - x + 3x^2y + 3xy^2 + y^3 - y
=x3+y3+3x2y+3xy2-x-y
=(x+y)(x2-xy+y2)+3xy(x+y)-(x+y)
=(x+y)(x2-xy+y2+3xy-1)
=(x+y)(x2+2xy+y2-1)
=(x+y)[(x+y)2-1]
=(x+y)(x+y-1)(x+y+1)
x^2 + 5x - 6
=x2-x+6x-6
=x.(x-1)+6.(x-1)
=(x-1)(x+6)
Ta có
a, x2-x-y2-y
=x2-y2-(x+y)
=(x-y)(x+y) - (x+y)
=(x+y)(x-y-1)
b, x2-2xy+y2-z2
=(x-y)2-z2
=(x-y-z)(x-y+z)
a) 7xy^2+5x^2y
= xy(7y+5x)
b) x^2+2xy+y^2-11x-11y
= (x^2+2xy+y^2)-(11x+11y)
=(x+y)^2-11(x+y)
=(x+y)(x+y-11)
a) 7xy2 + 5x2y
= xy ( 7y + 5x )
b) x2 + 2xy + y2 - 11x - 11y
= ( x2 + 2xy + y2 ) - ( 11x + 11y )
= ( x + y )2 - 11 ( x + y )
= ( x + y )( x + y - 11 )
\(4x^2-4y^2-4y-1\)
\(\Leftrightarrow4x^2-\left(2y+1\right)^2\)
\(\Leftrightarrow\left(2x-2y-1\right)\left(2x+2y+1\right)\)
P/s tham khảo nha
\(=4x^2-\left(4y^2+4y+1\right)\)
\(=4x^2-\left(4y^2+4y+1^2\right)\)
\(=\left(2x\right)^2-\left(2y+1\right)^2\)
\(=\left(2x-2y-1\right)\left(2x+2y+1\right)\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)+2-12\)
\(=\left(x^2+x\right)^2+3\left(x^2+x\right)-10\)
\(=\left(x^2+x+5\right)\left(x^2+x-2\right)\)
\(=\left(x^2+x+5\right)\left(x+2\right)\left(x-1\right)\)
\(x^2-x-4y^2-2y\)
\(=x^2-4y^2-\left(x+2y\right)\)
\(=\left(x-2y\right)\left(x+2y\right)-\left(x+2y\right)\)
\(=\left(x+2y\right)\left(x-2y-1\right)\)