Phân tích đa thức thành nhân tử(dạng đối xứng vòng quanh). 1)(a-b)(b-c)(a-c)+(a+b)(b+c)(a-c)+(a+b)(a+c)(c-b) 2)-(mab+n)(a-b)-(mbc+n)(b-c)-(mca+n)(c-a)

Question

Phân tích đa thức thành nhân tử(dạng đối xứng vòng quanh).
1)(a-b)(b-c)(a-c)+(a+b)(b+c)(a-c)+(a+b)(a+c)(c-b)
2)-(mab+n)(a-b)-(mbc+n)(b-c)-(mca+n)(c-a)
3)(aut+n)(u-t)+(auv+n)(v-u)+(atv+n)(t-v)
4)bc(a+d)(b-c)+ac(b+d)(c-a)+ab(c+d)(a-b)

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3 tuần 2021-08-13T13:41:53+00:00 2 Answers 0 views 0

1. 1)(a-b)(b-c)(a-c)+(a+b)(b+c)(a-c)+(a+b)(a+c)(c-b)

=a²b-ab²-a²c+abc-abc+b²c+ac²-bc²+ a²b+ab²+a²c+abc-abc-b²c-ac²-bc² -a²b-ab²+a²c+abc-abc-b²c+ac²+bc²

=a²b-ab²+a²c-b²c+ac²-bc²-abc+abc

=a(ab-b²+ac-bc)+c(ab-b²+ac-bc)

=[(ab-b²)+(ac-bc)](a+c)

=[b(a-b)+c(a-b)](a+c)

=(a-b)(b+c)(a+c)

2)-(mab+n)(a-b)-(mbc+n)(b-c)-(mca+n)(c-a)

=-ma²b+mab²-na+nb-mb²c+mbc²-nb+nc-mc²a+mca²-nc+na

=-m[a²b-ab²+b²c-bc²+c²a-ca²+abc-abc]

=-m[a(ab-b²+bc-ac)-c(ab+b²+bc-ac)]

=-m{(a-c)[ab-b²+bc-ac]}

=-m(a-b)(b+c)(a-c)

3)(aut+n)(u-t)+(auv+n)(v-u)+(atv+n)(t-v)

{Tương tự với (2)}

=a(t-v)(v+u)(u-t)

4)bc(a+d)(b-c)+ac(b+d)(c-a)+ab(c+d)(a-b)

=ab²c-abc²+b²cd-bc²d+abc²-a²bc+ac²d-a²cd+a²bc-ab²c+a²bd-ab²d

=b²cd-bc²d+ac²d-a²cd+a²bd-ab²d

=d[b²c-bc²+ac²-a²c+a²b-ab²]

=d[(b²c-bc²+ac²-abc)+(-a²c+a²b-ab²+abc)]

=d[c(b²-bc+ac-ab)-a(ac-ab+b²-bc)]

=d[(c-a)(b²-bc+ac-ab)]

=d(c-a)[(b²-bc)+(ac-ab)]

=d(c-a)[b(b-c)-a(b-c)]

=d(c-a)(b-c)(b-a)

2. 1)(a-b)(b-c)(a-c)+(a+b)(b+c)(a-c)+(a+b)(a+c)(c-b)

=a²b-ab²-a²c+abc-abc+b²c+ac²-bc²+ a²b+ab²+a²c+abc-abc-b²c-ac²-bc² -a²b-ab²+a²c+abc-abc-b²c+ac²+bc²

=a²b-ab²+a²c-b²c+ac²-bc²-abc+abc

=a(ab-b²+ac-bc)+c(ab-b²+ac-bc)

=[(ab-b²)+(ac-bc)](a+c)

=(a-b)(b+c)(a+c)

2)-(mab+n)(a-b)-(mbc+n)(b-c)-(mca+n)(c-a)

=-ma²b+mab²-na+nb-mb²c+mbc²-nb+nc-mc²a+mca²-nc+na

=-m[a²b-ab²+b²c-bc²+c²a-ca²+abc-abc]

=-m[a(ab-b²+bc-ac)-c(ab+b²+bc-ac)]

=-m(a-b)(b+c)(a-c)

3)(aut+n)(u-t)+(auv+n)(v-u)+(atv+n)(t-v)

=au²t-aut²+nu-nt+auv²-au²v+nv-nu+at²v-atv²+nt-nv

=au²t-aut²+auv²-au²v+at²v-atv²

=a(u²t-ut²+uv²-u²v+t²v-tv²)

=a(u²t-ut²+uv²-u²v+t²v-tv²+tuv-tuv)

=a[u(ut-t²+tv-uv)-v(-uv²-t²+tv+ut)]

=a(t-v)(v+u)(u-t)

4)bc(a+d)(b-c)+ac(b+d)(c-a)+ab(c+d)(a-b)

=ab²c-abc²+b²cd-bc²d+abc²-a²bc+ac²d-a²cd+a²bc-ab²c+a²bd-ab²d

=b²cd-bc²d+ac²d-a²cd+a²bd-ab²d

=d[b²c-bc²+ac²-a²c+a²b-ab²+abc-abc]

=d[c(b²-bc+ac-ab)-a(ac-ab+b²-bc)]

=d[(c-a)(b²-bc+ac-ab)]

=d(c-a)[b(b-c)-a(b-c)]

=d(c-a)(b-c)(b-a)

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