[Toán 8] Toán hay

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vipboycodon

Bài 2:

Ta có: $x+y+z = 0 \rightarrow z = -(x+y)$

Khi đó:

$x^5+y^5+z^5$

$= x^5+y^5-(x+y)^5$

$= x^5+y^5-(x^5+5x^4y+10x^3y^2+10x^2y^3+5xy^4+y^5)$

$= -5xy(x^3+2x^2y+2xy^2+y^3)$

$= -5xy(x+y)(x^2+xy+y^2)$

$= 5xyz(x^2+xy+y^2)$

$\rightarrow 2(x^5+y^5+z^5) = 5xyz(2x^2+2xy+2y^2) = 5xyz[x^2+y^2+(x+y)^2] = 5xyz(x^2+y^2+z^2)$
 
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