Đáp án:

 

Giải thích các bước giải:

$a)(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$

$=(6x+1)^2-2(6x+1)(6x-1)+(6x-1)^2$

$=[(6x+1)-(6x-1)]^2$

$=(6x+1-6x+1)^2$

$=2^2$

$=4$

$b) x(2x^2-3)-x^2(5x+1)+x^2$

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

$=-3x^3-3x$

$c) 3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

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

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

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

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

$=2^{32}-1$

$d) 3x(x-2)-5x(1-x)-8(x^2-3)$

$=3x^2-6x-5x+5x^2-8x^2+24$

$=-11x+24$

Đáp án:

a/ $4$

b/ $-3x^3-3x$

c/ $2^{32}-1$

d/ $-11x+24$

Giải thích các bước giải:

a/ $(6x+1)^2+(6x-1)^2-2(1+6x)(6x-1)$

$=[(6x+1)-(6x-1)]^2$

$=(6x+1-6x+1)^2$

$=2^2$

$=4$

b/ $x(2x^2-3)-x^2(5x+1)+x^2$

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

$=-3x^3-3x$

c/ $3(2^2+1)(2^4+1)(2^8+1)(2^{16}+1)$

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

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

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

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

$=2^{32}-1$

d/ $3x(x-2)-5x(1-x)-8(x^2-3)$

$=3x^2-6x-5x+5x^2-8x^2+24$

$=-11x+24$

Chúc bạn học tốt !!!