phan tich da thuc thanh nhan tu
a) 7xy^2+5x^2y
b) x^2+2xy+y^2-11x-11y
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(*)\(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(x-3\right)\left(3x-2\right)\)
(*)\(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-5\right)\left(x-1\right)\)
(*)\(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)^2-x^2=\left(x^2+1+x\right)\left(x^2+1-x\right)\)
(*)\(x^4-4x^2+3=x^4-x^2-3x^2+3=x^2\left(x^2-1\right)-3\left(x^2-1\right)=\left(x+1\right)\left(x-1\right)\left(x^2-3\right)\)
(*)\(6x^2+7xy+2y^2=6x^2+4xy+3xy+2y^2=2x\left(3x+2y\right)+y\left(3x+2y\right)=\left(2x+y\right)\left(3x+2y\right)\)
a, \(3x^2-11x+6=3x^2-2x-9x+6=x\left(3x-2\right)-3\left(3x-2\right)=\left(3x-2\right)\left(x-3\right)\)
b, \(x^2-6x+5=x^2-x-5x+5=x\left(x-1\right)-5\left(x-1\right)=\left(x-1\right)\left(x-5\right)\)
c, \(x^4+x^2+1=x^4+2x^2+1-x^2=\left(x^2+1\right)^2-x^2=\left(x^2+x+1\right)\left(x^2-x+1\right)\)
d, \(x^4-4x^2+3=x^4-4x^2+4-1=\left(x^2-2\right)^2-1=\left(x^2-1\right)\left(x^2-3\right)=\left(x+1\right)\left(x-1\right)\left(x^2-3\right)\)
e, \(6x^2+7xy+2y^2=6x^2+3xy+4xy+2y^2=3x\left(2x+y\right)+2y\left(2x+y\right)=\left(2x+y\right)\left(3x+2y\right)\)
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)
\(=3^2-\left(x-y\right)^2=\left[3-\left(x-y\right)\right]\left[3+\left(x-y\right)\right]=\left(3-x+y\right)\left(3+x-y\right)\)
\(9-x^2+2xy-y^2\)
\(=9-\left(x^2-2xy+y^2\right)\)
\(=3^2-\left(x-y\right)^2\)
\(=\left(3-x+y\right)\left(3-x-y\right)\)
x^2-2xy+y^2-9z^2
=(x-y)^2-9z^2
=(x-y)^2-(3z)^2
=(x-y-3z)(x-y+3z)
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 )