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\(3x^2-3y^2-2\left(x-y\right)^2\)
\(=3x^2-3y^2-2\left(x^2-2xy+y^2\right)\)
\(=3x^2-3y^2-2x^2+4xy-2y^2\)
\(=x^2+4xy-5y^2\)
\(=x^2+4xy+4y^2-9y^2\)
\(=\left(x+2y\right)^2-\left(3y\right)^2\)
\(=\left(x+2y-3y\right)\left(x+2y+3y\right)\)
\(=\left(x-y\right)\left(x+5y\right)\)
\(x^2-y^2+3x-3y\)
\(=\left(x^2-y^2\right)+\left(3x-3y\right)\)
\(=\left(x-y\right)\left(x+y\right)+3.\left(x-y\right)\)
\(=\left(x-y\right)\left(x+y+3\right)\)
\(x^2-y^2+4x+4\)
\(=\left(x^2+2.2x+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right)\left(x+2+y\right)\)
Tham khảo nhé~
x3+y(1-3x2)+x(3y2-1)-y3
= x3-3x2y+3xy2-y3+y-x
=(x-y)3 -(x-y)
=(x-y)(x2-2xy+y2-1)
cái chỗ kia giải thích dùm mìh đy : \(x^3-3x^2y+3xy^2-y^3+y-x\)
\(x^3+y\left(1-3x^2\right)+x\left(3y^2-1\right)-y^3\)
\(=x^3-3x^2y+3xy^2-y^3+y-x\)
\(=\left(x-y\right)^3-\left(x-y\right)\)
phân tích đa thức thành nhân tử cơ mà
=(x-y)3-(x-y)
=(x-y)[(x-y)2-1]
a) \(A=x^2-2xy+y^2+3x-3y-4\)
\(=\left(x-y\right)^2-1+3x-3y-3\)
\(=\left(x-y-1\right)\left(x-y+1\right)+3\left(x-y-1\right)\)
\(=\left(x-y-1\right)\left(x-y+1+3\right)\)
\(=\left(x-y-1\right)\left(x-y+4\right)\)
e) x2 -y2 + 3x - 3y = (x-y).(x+y) + 3.(x-y) = (x-y).(3+x+y)
g) x2 -y2 + 4x + 4 = (x-y).(x+y) + 4.(x+1) =
Câu g mình không giúp được . Xin lỗi bạn
e) x2 - y2 + 3x - 3y
= ( x - y ) ( x + y ) + 3 ( x - y )
= ( x - y ) ( x + y + 3 )
........
\(x^2-3x+xy-3y\)
\(=x\left(x+y\right)-3\left(x+y\right)\)
\(=\left(x+y\right)\left(x-3\right)\)
\(x^2-2xy+y^2-4=\left(x-y\right)^2-2^2=\left(x-y-2\right)\left(x-y+2\right)\)
\(x^2+x-y^2+y=\left(x-y\right)\left(x+y\right)+\left(x+y\right)=\left(x+y\right)\left(x-y+1\right)\)
x2 + y2 - 3x - 3y + 2xy
= ( x2 + 2xy + y2 ) - ( 3x + 3y )
= ( x + y )2 - 3( x + y )
= ( x + y )( x + y - 3 )
b) ( x2 - 4x )2 - 2( x - 2 )2 - 7
= ( x2 - 4x )2 - 2( x2 - 4x + 4 ) - 7 (*)
Đặt t = x2 - 4x
(*) <=> t2 - 2( t + 4 ) - 7
= t2 - 2t - 8 - 7
= t2 - 2t - 15
= t2 + 3t - 5t - 15
= t( t + 3 ) - 5( t + 3 )
= ( t + 3 )( t - 5 )
= ( x2 - 4x + 3 )( x2 - 4x - 5 )
= ( x2 - x - 3x + 3 )( x2 + x - 5x - 5 )
= [ x( x - 1 ) - 3( x - 1 ) ][ x( x + 1 ) - 5( x + 1 ) ]
= ( x - 1 )( x - 3 )( x + 1 )( x - 5 )
a) Ta có: \(x^2+y^2-3x-3y+2xy\)
\(=\left[\left(x^2+y^2+2xy\right)-2\left(x+y\right)+1\right]-\left(x+y+1\right)\)
\(=\left[\left(x+y\right)^2-2\left(x+y\right)+1\right]-\left(x+y+1\right)\)
\(=\left(x+y-1\right)^2-\left(x+y+1\right)\)
\(=\left(x+y-1\right)^2-\left(\sqrt{x+y+1}\right)^2\)
\(=\left(x+y-1+\sqrt{x+y+1}\right)\left(x+y-1-\sqrt{x+y+1}\right)\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)\)
\(=\left(x-y\right)\left(x+y+3\right)\)
\(=3\left(x-y\right)+\left(x-y\right)\left(x+y\right)=\left(x-y\right)\left(x+y+3\right)\)