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\(x^2+4x-y^2+4\)
\(=\left(x^2+2.x.2+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left(x+2-y\right).\left(x+2+y\right)\)
Tham khảo nhé~
\(x^2+4x-y^2+4\)
\(=x^2+4x+4-y^2\)
\(=\left(x^2+4x+4\right)-y^2\)
\(=\left(x^2+2x.2+2^2\right)-y^2\)
\(=\left(x+2\right)^2-y^2\)
\(=\left[\left(x+2\right)+y^2\right].\left[\left(x+2\right)-y^2\right]\)
\(=\left(x+2+y^2\right)\left(x+3-y^2\right)\)

Giải:
Đặt y = x^2 + x
Khi đó, đa thức trở thành:
xy^2 - 2y - 15
=xy^2 - 5y + 3y -15
= y(xy - 5) + 3(xy -5)
= (y+ 3)(xy -5)
Thay y vào, ta được:
(x^2 - x + 3)[x(x^2 - x) - 5]
=(x^2 - x + 3)(x^3 - x^2 - 5)
Sửa đề: \(\left(x^2+x\right)^2-2\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)^2-5\left(x^2+x\right)+3\left(x^2+x\right)-15\)
\(=\left(x^2+x\right)\left(x^2+x-5\right)+3\left(x^2+x-5\right)\)
\(=\left(x^2+x-5\right)\left(x^2+x+3\right)\)

Dùng hằng đẳng thức là xong
a, \(\left(x+y\right)^3-x^3-y^3=x^3+3x^2y+3xy^2+y^3-x^3-y^3\)
\(=3x^2y+3xy^2=3xy\left(x+y\right)\)
b, \(x^2+6xy+9y^2=\left(x+3y\right)^2\)

\(2x^2-x-15\)
\(=2x\left(x-3\right)+5\left(x-3\right)\)
\(=\left(2x+5\right)\left(x-3\right)\)

\(x^2-y^2+6x+9=\left(x+3\right)^2-y^2=\left(x+3+y\right)\left(x+3-y\right)\)
\(x^3+3x^2-9x-27=\left(x-3\right)\left(x^2+3x+9\right)+3x\left(x-3\right)=\left(x-3\right)\left(x^2+6x+9\right)=\left(x-3\right)\left(x+3\right)^2\)

\(e,x^2-y^2+2x+1=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2=\left(x+1-y\right)\left(x+1+y\right)\)
\(f,x^3+2x^2+2x+1=\left(x^3+1\right)+\left(2x^2+2x\right)\)
\(=\left(x+1\right)\left(x^2-x+1\right)+2x\left(x+1\right)\)
\(=\left(x+1\right)\left(x^2-x+1+2x\right)\)
\(=\left(x+1\right)\left(x^2+x+1\right)\)
\(x^2-y^2+2x+1\)
\(=\left(x^2+2x+1\right)-y^2\)
\(=\left(x+1\right)^2-y^2\)
\(=\left(x-y+1\right)\left(x+y+1\right)\)
hk tốt
^^

c, \(2x^2+x-3=x\left(2x-3\right)\)
d, \(6x^2-x-15=x\left(6x-15\right)\)
TK MIK NHA
\(2x^2+x-3=2x^2-2x+3x-3=2x\left(x-1\right)+3\left(x-1\right)=\left(x-1\right)\left(2x+3\right) \)

a) 16x2 - ( x2 + 4 )2
= ( 4x )2 - ( x2 + 4 )2
= [ 4x - ( x2 + 4 ) ][ 4x + ( x2 + 4 ) ]
= ( -x2 + 4x - 4 )( x2 + 4x + 4 )
= [ -( x2 - 4x + 4 ) ]( x + 2 )2
= [ -( x - 2 )2 ]( x + 2 )2
b) ( x + y )3 + ( x - y )3
= [ ( x + y ) + ( x - y ) ][ ( x + y )2 - ( x + y )( x - y ) + ( x - y )2 ]
= ( x + y + x - y )[ x2 + 2xy + y2 - ( x2 - y2 ) + x2 - 2xy + y2 ]
= 2x( 2x2 + 2y2 - x2 + y2
= 2x( x2 + 3y2 )
Cách 1: x 2 + 2xy - 15 y 2 = ( x 2 + 2xy + y 2 ) - 16 y 2
= x + y 2 - 4 y 2
= (x + y + 4y)(x + y – 4y)
= (x + 5y)(x – 3y).
Cách 2: x 2 + 2xy - 15 y 2 = x 2 + 5xy – 3xy - 15 y 2
= x(x + 5y) – 3y(x + 5y)
= (x – 3y)(x + 5y).