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a) \(x^2-x-y^2-y\)
\(=\left(x^2-y^2\right)-\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y\right)-\left(x+y\right)\)
\(=\left(x+y\right)\left(x-y-1\right)\)
a) \(^{x^2-x-y^2-y=\left(x^2-y^2\right)-\left(x+y\right)=\left(x-y\right)\left(x+y\right)-\left(x+y\right)=\left(x+y\right)\left(x-y-1\right)}\)
b)\(a^3-a^2x-ay=a\left(a^2-a.x-y\right)\)
c)\(5x^2-10xy+5y-20z^2=-20z^2+\left(5-10x\right)y+5x^2 \)
\(=-5\left(4z^2+2xy-y-x^2\right)\)
d)\(x^3-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3xy^2+3x^2y+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left[\left(x+y\right)^2-1\right]=\left(x+y\right)\left(x+y-1\right)\left(x+y+1\right)\)
a ) ( 3x2 + 3x + 2)2 - ( 3x2 + 3x - 2)2
=(3x2 + 3x + 2 + 3x2 + 3x - 2) [( 3x2 + 3x + 2) - ( 3x2 + 3x - 2) ]
=(6x2+6x)*4
=24x(x+1)
b ) ( xy+1)2 - ( x+y)2
=( xy+1 + x+y ) [( xy+1) - ( x+y)]
=[x(y+1)+(y+1)] [x(y-1) - (y-1)]
=(x+1)(y+1)(x-1)(y-1)
c ) ( x + y)3 - ( x - y)3
=[( x + y)-( x - y)] [( x + y)2 - ( x + y)( x - y) + ( x - y)2 ]
=2y( x2+2xy+y2 - x2+y2+ x2-2xy +y2 )
=2y(3y2+x2)
d ) 4( x2 - y2 ) - 8(x - ay) - 4(a2 - 1)
=4(-a2+2ay-y2+x2-2x+1)
=4[-(a-y)2+(x-1)2]
=-4(y-x-a+1)(y+x-a-1)
a) \(9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)
b) \(3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=\left(3x^4y^2+3xy^2\right)+\left(3x^3y^2+3y^2\right)\)
\(=3xy^2\left(x^3+1\right)+3y^2\left(x^3+1\right)\)
\(=\left(3xy^2+3y^2\right)\left(x^3+1\right)\)
\(=3y^2\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
c) \(\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2+x+y^2+y+1-xy\right)\)
Bài làm :
\(\text{a)}9\left(x+y-1\right)^2-4\left(2x+3y+1\right)^2\)
\(=\left(3x+3y-3\right)^2-\left(4x+6y+2\right)^2\)
\(=\left(3x+3y-3-4x-6y-2\right)\left(3x+3y-3+4x+6y+2\right)\)
\(=\left(-x-3y-5\right)\left(7x+9y-1\right)\)
\(\text{b)}3x^4y^2+3x^3y^2+3xy^2+3y^2\)
\(=\left(3x^4y^2+3xy^2\right)+\left(3x^3y^2+3y^2\right)\)
\(=3xy^2\left(x^3+1\right)+3y^2\left(x^3+1\right)\)
\(=\left(3xy^2+3y^2\right)\left(x^3+1\right)\)
\(=3y^2\left(x+1\right)\left(x+1\right)\left(x^2-x+1\right)\)
\(=3y^2\left(x+1\right)^2\left(x^2-x+1\right)\)
\(\text{c)}\left(x+y\right)^3-1-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left[\left(x+y\right)^2+x+y+1\right]-3xy\left(x+y-1\right)\)
\(=\left(x+y-1\right)\left(x^2+2xy+y^2+x+y+1-3xy\right)\)
\(=\left(x+y-1\right)\left(x^2+x+y^2+y+1-xy\right)\)
\(d ) x^3+3x^2+3x+1-27z^3\)
\(=\left(x+1\right)^3-\left(3z\right)^3\)
\(=\left(x+1-3z\right)\left(x^2+2x+1+3xz+3z+9z^2\right)\)
x^2-5x-14
=x^2-7x+2x-14
=(x^2-7x)+(2x+14)
=x(x-7)+2(x-7)
=(x+2)(x-7)
\(3x^2y-6xy^2+3xy\)
\(=3xy\left(x-2y+1\right)\)
\(x^2-5x=0\)
\(\Leftrightarrow x\left(x-5\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x=0\\x-5=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=0\\x=5\end{cases}}\)
\(x\left(x-1\right)-3x+3=0\)
\(\Leftrightarrow x\left(x-1\right)-3\left(x-1\right)=0\)
\(\Leftrightarrow\left(x-3\right)\left(x-1\right)=0\)
\(\Leftrightarrow\orbr{\begin{cases}x-3=0\\x-1=0\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=3\\x=1\end{cases}}\)
Bài 1 :
\(3x^2y-6xy^2+3xy\)
\(=3xy\left(x-2y+1\right)\)
a)x2-xy+x-y
=x(x-y)+(x-y)
=(x+1)(x-y)
b)3x2-3xy-5x+5y
=3x(x-y)-5(x-y)
=(3x-5)(x-y)
a ) \(x^2-xy+x-y\).
\(=x\left(x-y\right)+\left(x-y\right)\)
\(=\left(x-y\right)\left(x+1\right).\)
b ) \(3x^2-3xy-5x+5y\)
\(=3x\left(x-y\right)-5\left(x-y\right)\)
\(=\left(x-y\right)\left(3x-5\right)\)
x3 - x + 3x2y + 3xy2 + y3 - y ( sửa -x3 -> x3 )
= ( x3 + 3x2y + 3xy2 + y3 ) - ( x + y )
= ( x + y )3 - ( x + y )
= ( x + y )[ ( x + y )2 - 1 ]
= ( x + y )( x + y - 1 )( x + y + 1 )
\(a,=\left(m-y\right)\left(m+y\right)+a\left(m+y\right)=\left(m+y\right)\left(m-y+a\right)\\ b,=3x\left(y-1\right)+\left(y-1\right)\left(y+1\right)=\left(y-1\right)\left(3x+y+1\right)\)
a: \(=\left(m-y\right)\left(m+y\right)+a\left(m+y\right)\)
\(=\left(m+y\right)\left(m-y+a\right)\)