Phân Tích Đa Thức Thành Nhân Tử

T

transformers123

a/ $x^8+x^7+1$
$=x^8+x^7+x^6-x^6+1$
$=x^6(x^2+x+1)-(x^3-1)(x^3+1)$
$=x^6(x^2+x+1)-(x-1)(x^2+x+1)(x^3+1)$
$=(x^2+x+1)(x^6-x^4+x^3-x+1)$
b/ thêm bớt $x^6$ rồi làm tương tự=))
 
K

kenhaui

Bài 3:
$e$, $x^3+y^3+z^3-3xyz$ =$(x+y)^3-3x^2y-3xy^2+z^3-3xyz$
$=[(x+y)^3+z^3]-3xy(x+y+z)$
$= (x+y+z) [(x+y)^2-z(x+y)+z^2]-3xy(x+y+z)$
$= (x+y+z) (x^2+2xy+y^2-xz-yz+z^2-3xy)$
$= (x+y+z)(x^2+y^2+z^2-xy-yz-zx)$[/SIZE][/SIZE]
 
Last edited by a moderator:
K

kenhaui


$f$ ,$(x+y+z)^3-x^3-y^3-z^3=[(x+y)+z]^3-x^3-y^3-z^3$
$= (x+y)^3+z^3+3z(x+y)(x+y+z)-x^3-y^3-z^3$
$=x^3+y^3+3xy(x+y)+z^3+3(x+y)(x+y+z)-x^3-y^3-z^3$
$=3(x+y)(xy+xz+yz+z^2)$
$=3(x+y)[x(y+z)+z(y+z)]$
$=3(x+y)(y+z)(z+x) $
 
V

vipboycodon

d) $(x+1)(x+2)(x+3)(x+4)+1$
= $[(x+1)(x+4)][(x+2)(x+3)]+1$
= $(x^2+5x+4)(x^2+5x+6)+1$ (*)
Đặt $x^2+5x+4 = t$
=> (*) = $t(t+2)+1 = (t+1)^2 = (x^2+5x+5)^2$
 
You must log in or register to reply here.
Top Bottom