Chứng minh 2 đẳng thức bằng nhau

K

kiev

Ta có x^8 - y^8 = (x^4) ^2 - (y^4)^2
= (x^4 + y^4)(x^4 - y^4)
=(x^4 + y^4)(x^2 + y^2)(x^2 - y^2 )
= (x^4 + y^4)(x^2 + y^2)(x + y )(x - y )
=(x^4 + y^4)(x^2 + y^2)(x + y )(y +1-y)
=(x^4 + y^4)(x^2 + y^2)(x + y )

Chúc bạn thành công
 
I

icy_tears

$x = y + 1$

\Leftrightarrow $x - y = 1$


Ta có:

$(x + y)(x^2 + y^2)(x^4 + y^4)$

$= (x - y)(x + y)(x^2 + y^2)(x^4 + y^4)$

$= (x^2 - y^2)(x^2 + y^2)(x^4 + y^4)$

$= (x^4 - y^4)(x^4 + y^4)$

$= x^8 - y^8$

\Rightarrow ĐPCM
 
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