a: \(=\left(a+b+c\right)\cdot\dfrac{\left(2a^2+2b^2+2c^2-2ab-2bc-2ca\right)}{2}\)
\(=\dfrac{1}{2}\left(a+b+c\right)\left[\left(a-b\right)^2+\left(b-c\right)^2+\left(a-c\right)^2\right]\)
\(=a^3+b^3+c^3-3abc\)
b: \(=12^{2n-1}-3x^n-12^{2n-1}+2x^{n+1}\)
\(=-3x^n-2x^{n+1}\)
Cho a2+b2 +c2 -ab-ac-bc=0
Chứng minh a=b=c
(1) (a+b−c)2=a2+b2+c2+2ab−2bc−2ac(a+b−c)2=a2+b2+c2+2ab−2bc−2ac
(3) a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b)
(5) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
(7) (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
(11) ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
(1) (a+b−c)2=a2+b2+c2+2ab−2bc−2ac(a+b−c)2=a2+b2+c2+2ab−2bc−2ac
(3) a3+b3=(a+b)3−3ab(a+b)a3+b3=(a+b)3−3ab(a+b)
(5) (a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)(a+b+c)3=a3+b3+c3+3(a+b)(b+c)(c+a)
(7) (a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)(a−b)3+(b−c)3+(c−a)3=3(a−b)(b−c)(c−a)
(9) (a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc(a+b)(b+c)(c+a)=(a+b+c)(ab+bc+ca)−abc
(11) ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3ab3+bc3+ca3−a3b−b3c−c3a=(a+b+c)[(a−b)3+(b−c)3+(c−a)3]3
Phân tích các đa thức sau thành nhân tử:
a) (a+b+c)2-(ab+bc-ca)(a+b+c)+(a2+b+c)
b)a(a-b)(a+b)-(a+b)(a2-ab+b2)
c) 2x(x-1)-x(1-x)2-(1-x)3
d)xm+2 - xm
H·y viÕt c¸c biÓu thøc díi d¹ng tæng cña ba b×nh phong;
(a + b + c)2 + a2 + b2 + c2
mình ko biết tổng ba bình phong là gì
cho a,b,c là các số thực khác 0 thỏa mãn \(\frac{1}{a}+\frac{1}{b}+\frac{1}{c}=0\). tính A = \(\frac{bc}{a^2}+\frac{ca}{b^2}+\frac{ab}{c^2}\)
cho ab+bc+ca=1 chứng minh (a^2+1)(b^2+1)(c^2+1)=(a+b)^2(b+c)^2(c+a)^2
Bài 8: a)Chứng minh rằng ( a + b + c)3- a3 – b3 – c3 = 3( a +b)(b +c)( c+ a)
b)a3 +b3 +c3 – 3abc = ( a + b + c)( a2 +b2 + c2)
P = ( a+1) ^2+(b+1) ^2+(c+1) ^2+2( ab+bc+ca)
Q= ( a+b+c+1) ^2
Tính P-Q
