\(x^4+y^4+\left(x+y\right)^4\)

\(=x^4+y^4+\left(x^2+2xy+y^2\right)^2\)

\(=x^4+y^4+x^4+6x^2y^2+y^4+4x^3y+4xy^3\)

\(=2.\left(x^2+y^2\right)^2+4xy\left(x^2+y^2\right)+2x^2y^2\)

\(=2.\left(x^2+y^2\right)\left(x^2+y^2+2xy\right)+2x^2y^2\)

\(=2.\left[\left(x^2+y^2\right)\left(x+y\right)^2+x^2y^2\right]\)

Sai thì thôi nhé~

       \(x^4+y^4+\left(x+y\right)^4\)

\(=x^4+y^4+x^4+4x^3y+6x^2y^2+4xy^3+y^4\)

\(=2x^4+4x^3y+6x^2y^2+4xy^3+2y^4\)

\(=2\left(x^4+2x^3y+3x^2y^2+2xy^3+y^4\right)\)

\(=2\left[\left(x^4+2x^3y+x^2y^2\right)+2\left(x^2+xy\right)y^2+y^4\right]\)

\(=2\left[\left(x^2+xy\right)^2+2\left(x^2+xy\right)y^2+\left(y^2\right)^2\right]\)

\(=2\left(x^2+xy+y^2\right)^2\)

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 )

nâng cao phát triển toán 8 tập 1 mình ngại viết nên bạn vào đó xem nhé

Xét riêng (x + y)^4 = [(x + y)^2]^2 = [x^2+2xy+y^2]^2 = x^4 +4x^2y^2 + y^4 + 4x^3y + 2x^2y^2+4xy^3
Vậy (x + y)^4 +x^4 + y^4 = x^4 +4x^2y^2 + y^4 + 4x^3y + 2x^2y^2+4xy^3+ x^4 + y^4 
= 2x^4 + 2y^4 + 6x^2y^2 + 4x^3y + 4xy^3 
= 2(x^4 + y^4 + 3x^2y^2 +2 x^3y + 2xy^3) 
= 2(x^4 + y^4 + x^2y^2 + 2x^3y + 2xy^3 + 2x^2y^2) 
= 2(x^2 + xy + y^2)^2

Ây za,mik ko bt có đúng ko nhưng mik thử làm nhé.

Đặt \(x^4+y^4+z^4=a;x^2+y^2+z^2=b;x+y+z=c\)

\(\Rightarrow M=2a-b^2-2bc^2+c^4\)

\(M=2a-2b^2+b^2-2bc^2+c^4\)

\(M=2\left(a-b^2\right)+\left(b-c^2\right)^2\)

Mà:

\(a-b^2=-2\left(x^2y^2+y^2z^2+z^2x^2\right)\)

\(b-c^2=-2\left(xy+yz+zx\right)\)

Khi đó:

\(M=-4\left(x^2y^2+y^2z^2+z^2x^2\right)+4\left(xy+yz+zx\right)^2\)

\(M=-4x^2y^2-4y^2z^2-4z^2x^2+4x^2y^2++4y^2z^2+4z^2x^2+4z^2x^2+8x^2yz+8xy^2z+8xyz^2\)

\(M=8xyz\left(x+y+z\right)\)

\(4\left(1+x\right)\left(1+y\right)\left(1+x+y\right)-3x^2y^2=4\left(1+x+y+xy\right)\left(1+x+y\right)-3x^2y^2\)

\(=4\left(1+x+y\right)^2+4xy\left(1+x+y\right)+x^2y^2-4x^2y^2\)

\(=\left[2\left(1+x+y\right)+xy\right]^2-\left(2xy\right)^2=\left(2+2x+2y+xy-2xy\right)\left(2+2x+2y+xy+2xy\right)\)

\(=\left(2+2x+2y-xy\right)\left(2+2x+2y+3xy\right)\)

giúp mình câu khác được ko? câu này mình biết làm òi

rygghgjgfhgfhgfhgfnb45 - u6 

\(x^2\left(x+4\right)^2-\left(x+4\right)^2-\left(x^2-1\right)\)

\(=\left(x+4\right)^2\left(x^2-1\right)-\left(x^2-1\right)\)

\(=\left(x^2-1\right)\left[\left(x+4\right)^2-1\right]\)

\(=\left(x-1\right)\left(x+1\right)\left(x+4+1\right)\left(x+4-1\right)\)

\(=\left(x-1\right)\left(x+1\right)\left(x+5\right)\left(x-3\right)\)

=.= hok tốt!!