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H

huynhbachkhoa23

Bài 1: $(a+b+c)(ab+bc+ca)-abc=(a+b)(ab+bc+ca)+c(ab+bc+ca)-abc$

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

thopeo_kool

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

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

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

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

$= 3(a + b)[(b + c)(c + a) - (b - c)(a - c)] - 3(a - b)[(b + c)(a - c) + (a + c)(c - b)]$

$= 3(a + b)[bc + ba + c^2 + ca - (ba - bc - ca + c^2)] - 3(a - b)[ba - bc + ca - c^2 + (ac - ab + c^2 - cb)]$

$= 3(a + b)(bc + ba + c^2 + ac - ba + bc + ca - c^2) - 3(a - b)(ba - bc + ca - c^2 + ac - ab + c^2 - cb)$

$= 6c(a + b)^2 - 6c(a -b)^2 = 6c[(a + b) - (a - b)].[(a + b) + (a - b)] = 6c.2b.2a = 24abc$
 
V

vipboycodon

3) $(x+2)^4+(x-2)^4-32$
= $x^4+8x^3+24x^2+32x+16+x^4-8x^3+24x^2-32x+16-32$
= $2x^4+48x^2$
= $2x^2(x^2+24)$

4) $(x+1)^4+(x-1)^4-2$
= $x^4+4x^3+6x^2+4x+1+x^4-4x^3+6x^2-4x+1-2$
= $2x^4+12x^2$
= $2x^2(x^2+6)$
 
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