[Toán 8] Chứng minh

C

chonhoi110

Ta có: $x^3+y^3=(x+y)^3-3xy(x+y)$
=> $xy=\dfrac{1-a}{3}$
Do đó: $x^2 + y^2 = (x + y)^2 – 2xy = 1 - 2.\dfrac{1-a}{3} = \dfrac{1+2a}{3}$
Vì $x^5 + y^5 = (x^3 + y^3)( x^2 + y^2) – x^2y^2(x + y)$
=> $b = a.\dfrac{(1+2a)}{3}-\dfrac{(1-a)^2}{9}$ hay $5a (a + 1) = 9b + 1$
=> đpcm
 
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