Toán 8 - không nâng cao

N

noinhobinhyen

a.

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

b.

$x^4-5x^2+4 = (x-2)(x-1)(x+1)(x+2)$

c.

$(x+y+z)^3-x^3-y^3-z^3=3(x+y)(y+z)(z+x)$
 
D

dapsieuchuan

Làm được câu a với câu b thôi còn câu c hơi dài

a) x3 - 3x2 – 4x + 12
= (x3 - 3x2) – (4x – 12)
= x2(x-3) – 4(x-3)
=(x-3)(x2-4)
=(x-3)(x-2)(x+2)
b) x4 – 5x2 + 4
=x4 – x2 - 4x2 ­­­ + 4
=(x4 – 4x2) – (x2 – 4)
=x2(x2-4) – (x2 – 4)
=(x2 – 4)(x2 – 1)
=(x-2)(x+2)(x-1)(x+1)


Nhớ thì thanksss nha
 
J

jin_luv_rain

nè pn

a) x^2*(x-3)-4(x-3)
= (x-3)(x^2-4)
= (x-3)(x^2-2^2)
= (x-3)(x-2)(x+2)

b) x^4-5x^2+4
=x^4-4x^2-x^2+4
=x^2(x^2-4)-(x^2-4)
=(x^2-4)(x^2-1)
=(x-2)(x+2)(x-1)(x+1)

c) (x+y+z)^3-x^3-y^3-z^3
lam` tươg tự nhé pn
 
C

chipcoi_no.love

[tex](x+y+z)^3-x^3-y^3-z^3[/tex]

[tex]=(x+y+z-x)[(x+y+z)^2+(x+y+z)x+x^2]-(y+z)(y^2-yz+z^2)[/tex]

[tex]=(y+z)[(x+y+z)^2+(x+y+z)x+x^2-(y^2-yz+z^2)][/tex]

[tex]=(y+z)[(x^2+y^2+z^2+2xy+2yz+2xz)+x^2+xy+xz+x^2-y^2+yz-z^2][/tex]

[tex]=(y+z)(3x^2+3xy+3yz+3xz)[/tex]

[tex]=3(y+z)[(x^2+xy)+(yz+xz)][/tex]

[tex]=3(y+z)[x(x+y)+z(y+x)][/tex]

[tex]=3(y+z)(x+y)(x+z)[/tex]
 
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