Giải thích các bước giải:
5/. $x^4y^4+4$
= $(x^2y^2)^2+2.x^2y^2.2-4x^2y^2+4$
= $[(x^2y^2)^2+2.x^2y^2.2+2^2]-4x^2y^2$
= $(x^2y^2+2)^2-(2xy)^2)$
= $[(x^2y^2+2)-2xy][(x^2y^2+2)+2xy]$
= $(x^2y^2+-2xy)(x^2y^2+2+2xy)$
= $(x^2y^2-2xy+2)(x^2y^2+2xy+2)$
6/. $x^4y^4+64$
= $x^4y^4+16x^2y^2-16x^2y^2+64$
= $(x^4y^4+16x^2y^2+64)-16x^2y^2$
= $(x^2y^2)^2+2.x^2y^2.8+8^2)-16x^2y^2$
= $(x^2y^2+8)^2-(4xy)^2$
= $[(x^2y^2+8)-4xy][(x^2y^2+8)+4xy]$
= $(x^2y^2+8-4xy)((x^2y^2+8+4xy)$
= $(x^2y^2-4xy+8)((x^2y^2+4xy+8)$
7/. $4x^4+1$
= $4x^4+4x^2-4x^2+1$
= $(4x^4+4x^2+1)-4x^2$
= $[(2x^2)^2+2.x^2.2+1^2)-4x^2$
= $(2x^2+1)^2)-(2x)^2$
= $(2x^2+1-2x)(2x^2+1+2x)$
= $(2x^2-2x+1)(2x^2+2x+1)$
8/. $32x^4+1$
Mình không biết làm
9/. $x^4+4y^4$
= $(x^2)^2+(2y^2)^2$
= $(x^2)^2+4x^2y^2-4x^2y^2+(2y^2)^2$
= $[(x^2)^2+4x^2y^2+(2y^2)^2]-4x^2y^2$
= $[(x^2)^2+2.x^2.2y^2+(2y^2)^2]-(2xy)^2$
= $[(x^2+2y^2)^2]-(2xy)^2$
= $(x^2+2y^2-2xy)(x^2+2y^2+2xy)$
10/. $x^7+x^2+1$
= $x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x+x^2+1$
= $(x^7+x^6+x^5)-(x^6+x^5+x^4)+(x^4+x^3+x^2)-(x^3+x^2+x)+(x^2-x+1)$
= $(x^5(x^2+x+1)-x^4(x^2+x+1)+x^2(x^2+x+1)-x(x^2+x+1)+(x^2-x+1)$
= $(x^2+x+1)(x^5-x^4+x^2-x+1)$
11/. $x^8+x+1$
= $x^8+x^7-x^7+x^6-x^6+x^5-x^5+x^4-x^4+x^3-x^3+x^2-x^2+x+1$
= $(x^8+x^7+x^6)-(x^7+x^6+x^5)+(x^5+x^4+x^3)-(x^4+x^3+x^2)+(x^2+x+1)$
= $x^6(x^2+x+1)-x^5(x^2+x+1)+x^3(x^2+x+1)-x^2(x^2+x+1)+(x^2+x+1)$
= $(x^2+x+1)(x^6-x^5+x^3-x^1-1)$
12/. $x^8+x^7+1$
= $x^8+x^7+1$
= $x^8+x^7+x^6-x^6+1$
= $(x^8+x^7+x^6)-(x^6-1)$
= $x^6(x^2+x+1)-[(x^3)^2-1]$
= $x^6(x^2+x+1)-(x^3-1)(x^3+1)$
= $x^6(x^2+x+1)-(x-1)(x^2+x+1)(x^3+1)$
= $(x^2+x+1)(x^6-(x-1)(x^3+1)$
= $(x^2+x+1)(x^6-(x^4+x-x^3-1)$
= $(x^2+x+1)(x^6-x^4-x+x^3+1)$
13/. $x^8+3x^4+1$
Bạn xem lại đề
14/. $x^10+x^5+1$
= $x^10+x^9-x^9+x^8-x^8+x^7-x^7+x^6-x^6+x^5+x^4-x^4+x^3-x^3+x^2-x^2+x-x+1$
= $(x^10+x^9+x^8)-(x^9+x^8+x^7)+(x^7+x^6+x^5)-(x^6+x^5+x^4)+(x^5+x^4+x^3)-(x^3+x^2+x)+(x^2+x+1)$
= $x^8(x^2+x+1)-x^7(x^2+x+1)+x^5(x^2+x+1)-x^4(x^2+x+1)+x^3(x^2+x+1)-x(x^2+x+1)+(x^2+x+1)$
= $(x^2+x+1)(x^8-x^7+x^5-x^4+x^3-x+1)$
Chúc bạn học tốt
