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18 tháng 8 2020

\(\left(a-b\right).\left(b-c\right).\left(c-a\right)\)

\(=\left(ab-ac-b^2+bc\right).\left(c-a\right)\)

\(=abc-a^2b-ac^2+a^2c-b^2c+ab^2+bc^2-abc\)

\(=-a^2b-ac^2+a^2c-b^2c+ab^2+bc^2.\)

6 tháng 7 2016

=2*(c^2+b^2+a^2)

18 tháng 12 2022

lỗi

13 tháng 8 2015

\(\frac{1}{\left(a-b\right)\cdot\left(b-c\right)}-\frac{1}{\left(a-c\right)\cdot\left(b-c\right)}-\frac{1}{\left(a-b\right)\cdot\left(a-c\right)}\)

\(=\frac{a-c-\left(a-b\right)-\left(b-c\right)}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{a-c-a+b-b+c}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}\)

\(=\frac{0}{\left(a-b\right)\left(b-c\right)\left(a-c\right)}=0\)

13 tháng 8 2015

\(\frac{1}{\left(a-b\right).\left(b-c\right)}-\frac{1}{\left(a-c\right).\left(b-c\right)}-\frac{1}{\left(a-b\right).\left(a-c\right)}\)

=\(\frac{a-c}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{a-b}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}-\frac{b-c}{\left(a-b\right).\left(b-c\right).\left(c-a\right)}\)

=\(\frac{\left(a-c\right)-\left(a-b\right)-\left(b-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)

=\(\frac{a-c-a+b-b+c}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)

=\(\frac{\left(a-a\right)+\left(b-b\right)+\left(c-c\right)}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)

=\(\frac{0}{\left(a-b\right).\left(b-c\right).\left(a-c\right)}\)

=0

7 tháng 5 2018

a) x 3   –   3 x 2  + 3x – 1;

b) – x 4   +   7 x 3   –   11 x 2  + 6x – 5;

c) c 3   +   2 c 2  – 5c – 6.

a) (x - 2)(x2 - 2x + 4)(x - 2)( x2 + 2x + 4)

= (x - 2)2(x - 2)2(x + 2)2

= (x - 2)4(x + 2)2

b) (a + b + c)3 - (b + c - a)3 - (a - b + c)3 - (a + b - c)3

Đặt a+b-c=x, c+a-b=y, b+c-a=z

=>x+y+z=a+b-c+c+a-b+b+c-a=a+b+c

Ta có hằng đẳng thức:

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

=>(a+b+c)^3-(b+c-a)^3-(a+c-b)^3-(a+b-c)^3=(x+y+z)^3-x^3-y^3-z^3

=3(x+y)(x+z)(y+z)

=3(a+b-c+c+a-b)(c+a-b+b+c-a)(b+c-a+a+b-c)

=3.2a.2b.2c

=24abc


c) (a + b)3 + (b + c)3 + (c + a)3 - 3(a + b)(b + c)(c + a)

Đặt x = a+b; y = b+c; z = c+a ta có:

x3+y3+z3−3xyz

= (x+y)3−3xy(x−y)+z3−3xyz

=[(x+y)3+z3]−3xy(x+y+z)

=(x+y+z)3−3z(x+y)(x+y+z)−3xy(x−y−z)

=(x+y+z)[(x+y+z)2−3z(x+y)−3xy]

=(x+y+z)(x2+y2+z2+2xy+2xz+2yz−3xz−3yz−3xy)

=(x+y+z)(x2+y2+z2−xy−yz−yx)

Thay vào ta có:

(a+b+b+c+c+a)[(a+b)2+(b+c)2+(c+a)2−(a+b)(b+c)−(b+c)(c+a)−(c+a)(a+b)]

=(2a+2b+2c)(a2−ab−ac+b2−bc+c2)

=2(a+b+c)(a2−ab−ac+b2−bc+c2)

7 tháng 6 2019

a)\(\left(x-2\right)\left(x^2-2x+4\right)\left(x-2\right)\left(x^2+2x-4\right)\)

\(=\left(x-2\right)^2\left(x^2-2x+4\right)\left(x^2+2x-4\right)\)

\(=\left(x-2\right)^2\left(x^4+4x^2+16\right)\)

\(=x^6-4x^5+8x^4-16x^3+32x^2-64x+64\)