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\(3y^3+6xy^2+3x^2y=3y\left(y^2+2xy+x^2\right)=3y\left(x+y\right)^2\)
\(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x^2-4\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
\(x^3+3x^2-3x-1=\left(x-1\right)\left(x^2+x+1\right)+3x\left(x-1\right)=\left(x-1\right)\left(x^2+x+1+3x\right)\)
\(=\left(x-1\right)\left(x^2+4x+1\right)\)
Tham khảo nhé~
A/ \(16x-5x^2-3=\left(15x-3\right)-\left(5x^2-x\right)=3\left(5x-1\right)-x\left(5x-1\right)=\left(5x-1\right)\left(3-x\right)\)
B/ \(x^3-3x^2+1-3x=\left(x^3-4x^2+x\right)+\left(x^2-4x+1\right)=x\left(x^2-4x+1\right)+\left(x^2-4x+1\right)\)
\(=\left(x+1\right)\left(x^2-4x+1\right)\)
C/ \(x^3-3x^2-4x+12=x^2\left(x-3\right)-4\left(x-3\right)=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
D/ \(\left(2x+1\right)^2-\left(x-1\right)^2=\left(2x+1-x+1\right)\left(2x+1+x-1\right)=3x\left(x+2\right)\)
a ) \(x^3-3x^2-4x+12\)
\(=x^2\left(x-3\right)-4\left(x-3\right)\)
\(=\left(x^2-4\right)\left(x-3\right)\)
\(=\left(x-3\right)\left(x-2\right)\left(x+2\right)\)
b ) \(x^4-5x^2+4\)
\(=x^4-4x^2-x^2+4\)
\(=x^2\left(x^2-4\right)-\left(x^2-4\right)\)
\(\Leftrightarrow\left(x^2-1\right)\left(x^2-4\right)\)
\(=\left(x-1\right)\left(x+1\right)\left(x-2\right)\left(x+2\right)\)
\(=\left(x-2\right)\left(x-1\right)\left(x+1\right)\left(x+2\right)\)
a) \(x^3-3x+1-3x^2=\left(x^3+1\right)-\left(3x^2+3x\right)=\left(x+1\right)\left(x^2-x+1\right)-3x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1-3x\right)=\left(x+1\right)\left(x^2-4x+1\right)\)
b) \(2x^2+4x+2-2y^2=2\left(x^2+2x+1-y^2\right)=2\left[\left(x+1\right)^2-y^2\right]=2\left(x+1+y\right)\left(x+1-y\right)\)
a , 3x2 + 3y2 - 6xy - 12
= 3 ( x2 + y2 - 2xy - 4 )
= 3 ( x - y )2 - 22
= 3 ( x - y + 2 ) ( x - y - 2 )
a) \(x^2+4x+3=\left(x^2+4x+4\right)-1=\left(x+2\right)^2-1^2=\left(x+1\right)\left(x+3\right)\) (mình sửa lại)
b) \(x^2+8x-9=\left(x^2+8x+16\right)-25=\left(x+4\right)^2-5^2=\left(x-1\right)\left(x+9\right)\)
c) \(3x^2+6x-9=3\left[\left(x^2+2x+1\right)-4\right]=3\left[\left(x+1\right)^2-2^2\right]=3\left(x-1\right)\left(x+3\right)\)
d) \(2x^2+x-3=2x^2-4x+2+5x-5=2\left(x^2-2x+1\right)+5\left(x-1\right)=2\left(x-1\right)^2+5\left(x-1\right)=\left(x-1\right)\left(2x+3\right)\)
Câu 2 nha
\(a,x^4+2x^3+x^2\)
\(=x^2\left(x^2+2x+1\right)\)
\(=x^2\left(x+1\right)^2\)
\(c,x^2-x+3x^2y+3xy^2+y^3-y\)
\(=\left(x^3+3x^2y+3xy^2+y^3\right)-\left(x+y\right)\)
\(=\left(x+y\right)^3-\left(x+y\right)\)
\(=\left(x+y\right)\left(x^2+2xy+y^2-1\right)\)
a ) x3 - 2x2 + x
= x3 - x2 - x2 + x
= ( x3 - x2 ) - ( x2 - x )
= x2 . ( x - 1 ) - x . ( x - 1 )
= ( x - 1 )( x2 - x )
b ) x3 - 3x2 - 4x + 12
= ( x3 - 3x2 ) - ( 4x - 12 )
= x2 . ( x - 3 ) - 4 . ( x - 3 )
= ( x - 3 )( x2 - 4 )
= ( x - 3 )( x - 2 )( x + 2 )