\(VT=-a-b-c-1+b-c-a+c+1\)

        \(=-c-2a=VP\)

VẬY ....

Ta có

\(VT:-\left(a+b+c+1\right)+\left(b-c\right)-\left(a-c-1\right)\)

\(\Leftrightarrow-a-b-c-1+b-c-a+c+1\)

\(\Leftrightarrow-2a-c=VT\)

\(\Rightarrow-\left(a+b+c+1\right)+\left(b-c\right)-\left(a-c-1\right)=-c-2a\left(đpcm\right)\)

Bài này bạn chỉ cần phá ngoặc ra và tính là đc

Ta có : \(VT=\frac{\left(2a+2b-c\right)^2+\left(2b+2c-a\right)^2+\left(2c+2a-b\right)^2}{9}\)

\(=\frac{4a^2+4b^2+8ab+c^2-4ac-4ab+4b^2+4c^2+8bc+a^2-4ba-4bc+4c^2+4a^2+8ac+b^2-4bc-4ab}{9}\)\(=\frac{9\left(a^2+b^2+c^2\right)}{9}=a^2+b^2+c^2=VP\)

Vậy ta có đẳng thức: 

\(\left(\frac{2a+2b-c}{3}\right)^2+\left(\frac{2b+2c-a}{3}\right)^2+\left(\frac{2c+2a-b}{3}\right)^2=a^2+b^2+c^2\)

\(\left(a+b+c\right)^2+a^2+b^2+c^2=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)

VT : (a + b + c)² + a² + b² + c²

= a² + b² + c² + 2ab +2bc + 2ac + a² + b² + c²

= ( a² + 2ab + b² ) + (b² + 2bc + c²) + ( a² + 2ac + c²)

= (a + b)² + (b + c)² + (a + c)² = VP

Vậy \(\left(a+b+c\right)^2+a^2+b^2+c^2=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2\)(đpcm)

Để chứng minh hằng đẳng thức a^3 + b^3 + c^3 + 3(a+b)(b+c)(c+a) = (a+b+c)^3, ta sẽ sử dụng công thức khai triển đa thức.

Theo công thức khai triển đa thức, ta có:

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

Vậy, hằng đẳng thức được chứng minh.

(a+b+c)^3=((a+b)+c)^3=(a+b)^3+c^3+3(a+b)c(a+b+c)
=a^3+b^3+3ab(a+b)+c^3+3(a+b)c(a+b+c)
=a^3+b^3+c^3+3(a+b)(ab+c(a+b+c))
=a^3+b^3+c^3+3(a+b)(ab+ac+bc+c^2)
=a^3+b^3+c^3+3(a+b)(a+c)(b+c)

Ta có :\(\frac{b+c}{\left(a-b\right)\left(a-c\right)}+\frac{c+a}{\left(b-c\right)\left(b-a\right)}+\frac{a+b}{\left(c-a\right)\left(c-b\right)}\)​

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

\(=\frac{b^2-c^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{c^2-a^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}+\frac{a^2-b^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}\)

\(=\frac{b^2-c^2+c^2-a^2+a^2-b^2}{\left(a-b\right)\left(a-c\right)\left(b-c\right)}=0\left(ĐPCM\right)\)

Có: Vế trái : (a - c)(b + d) - (a - d)(b + c) 

= ab + ad - bc - cd - ab - ac + bd + cd

= ad - bc - ac + bd

= ad - ac + bd + bc

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

= (a + b)(d - c) (= vế phải)

Vậy đpcm

BĐVT có,

=ab+ad-bc-cd-ab-ac+bd+cd

=ad-ac-bc+bd

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

=(a+b)(d-c)=vế phải

suy ra đpcm

tik nha

\(\left(a-c\right)\left(b+d\right)-\left(a-d\right)\left(b+c\right)=ab+ad-cb-cd-\left(ab+ac-bd-cd\right)\)

\(=ab+ad-cb-cd-ab-ac+bd+cd=ad-cb-ac+bd=\left(ad-ac\right)+\left(bd-bc\right)\)

\(=a\left(d-c\right)+b\left(d-c\right)=\left(a+b\right)\left(d-c\right)\)