[Toán 8] Hằng đẳng thức

I

icy_tears

$12(5^2 + 1)(5^4 + 1)(5^8 + 1)(5^{16} + 1)$

= $\frac{1}{2} (5^2 - 1)(5^2 + 1)(5^4 + 1)(5^8 + 1)(5^{16} + 1)$

= $\frac{1}{2} (5^4 - 1)(5^4 + 1)(5^8 + 1)(5^{16} + 1)$

= $\frac{1}{2} (5^8 - 1)(5^8 + 1)(5^{16} + 1)$

= $\frac{1}{2} (5^{16} - 1)(5^{16} + 1)$

= $\frac{1}{2} (5^{32} - 1)$
 
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I

icy_tears

$(a + b + c)^3$

$= (a + b)^3 + c^3 + 3(a + b)^2c + 3(a + b)c^2$

$= a^3 +b^3 + 3a^2b + 3ab^2 + c^3 + 3(a + b)^2c + 3(a + b)c^2$

$= a^3 +b^3 + 3ab(a + b) + c^3 + 3(a + b)^2c + 3(a + b)c^2$

$= a^3 + b^3 + c^3 + 3(a + b)[ab + c(a + b) + c^2)]$

$= a^3 + b^3 + c^3 + 3(a + b)(ab + bc + ca + c^2)$

$= a^3 + b^3 + c^3 + 3(a + b)(b + c)(c + a)$

(ĐPCM)
 
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T

thaiha_98

Bài 3.5/8 SBT: Chứng minh:
(a+b+c)^3 = a^3+b^3+c^3 + 3(a+b)(a+c)(b+c):khi (76)
Ta có:
$(a+b+c)^3 = ((a+b)+c)^3$
$ = (a+b)^3 + c^3 + 3(a+b)^2c + 3(a+b)c^2$
$ = a^3 + b^3 + 3a^2b + 3ab^2 + c^3 + 3(a+b)^2c + 3(a+b)c^2$
$ = a^3 + b^3 + c^3 + 3ab(a+b) + 3(a+b)^2c + 3(a+b)c^2$
$ = a^3 + b^3 + c^3 + 3 (a+b) ( ab + c(a+b) + c^2))$
$ = a^3 + b^3 + c^3 + 3 (a+b) ( ab + ca + cb + c^2)$
$ = a^3 + b^3 + c^3 + 3 (a+b) ( a(b+c) + c (b+c) )$
$ = a^3 + b^3 + c^3 + 3 (a+b) (b+c) (a+c) $ (đpcm)
 
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