rút gọn biểu thức .

Giải thích các bước giải:

a.Ta có:

$(2x+3)^2-2(2x+3)(2x+5)+(2x+5)^2$

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

$=(-2)^2$

$=4$

b.Ta có:

$(x^2+x+1)(x^2-x+1)(x^2-1)$

$=(x^2+x+1)(x^2-x+1)\cdot (x-1)(x+1)$

$=(x-1)(x^2+x+1)\cdot (x+1)(x^2-x+1)$

$=(x^3-1)\cdot (x^3+1)$

$=(x^3)^2-1^2$

$=x^6-1$

c.Ta có:

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

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

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

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

$=2a^2$

d.Ta có:

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

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

$=4a^2-2(a+b+c)(a-b-c)+4a^2-2(b-c-a)(c-a-b)$

$=8a^2-2((a+b+c)(a-b-c)+(b-c-a)(c-a-b))$

$=8a^2-2((a+b+c)(a-(b+c))-(b-c-a)(-c+a+b))$

$=8a^2-2((a+(b+c))(a-(b+c))-((b-c)-a)((b-c)+a))$

$=8a^2-2(a^2-(b+c)^2-((b-c)^2-a^2))$

$=8a^2-2(a^2-(b+c)^2-(b-c)^2+a^2)$

$=8a^2-2(2a^2-(b^2+2bc+c^2)-(b^2-2bc+c^2))$

$=8a^2-2(2a^2-2b^2-2c^2)$

$=8a^2-4a^2+4b^2+4c^2$

$=4a^2+4b^2+4c^2$