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

=2a^2+2b^2+2c^2-2bc-2ab-2ac

=a^2-2ac+c^2+a^2-2ab+b^2+b^2-2bc+c^2

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

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

\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)

\(=a^2+2ab+b^2+b^2+2bc+c^2+c^2+2ca+a^2\)

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

a,(a+b+c)^2+a^2+b^2+c^2

=a^2+b^2+c^2+2ab+2ac+2bc+a^2+b^2+c^2

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

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

b,(2a-b)(c-b)+2(b-a)(c-a)+2(b-c)(a-c)

=2(a-b)(c-b-c+a)+2(b-c)(c-a)

=2(a-b)(a-b)+2(b-c)(c-a)

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

=2(a^2-2ab+b^2)+(ab-bc-ca+c^2)

=2(a^2+b^2+c^2-ab-bc-ca)

=(a^2-2ab+b^2)+(b^2-2bc+c^2)+(c^2-2ca+a^2)

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

chúc bạn học tốt!!!

mơn nhìu

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

\(=\left(a-c\right)^2-\left(a-b\right)^2-\left(b-c\right)^2\) tương tự thì

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

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

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

Ta có: \(\left(a^2+b^2\right)\left(c^2+d^2\right)\)

\(=a^2c^2+a^2d^2+b^2c^2+b^2d^2\)

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

\(=\left(ac+bd\right)^2+\left(ad-bc\right)^2\)

=> đpcm

1) a) a^2+b^2=ab+ba

<=> a^2+b^2-2ab=0

<=> (a-b)^2=0

<=> a-b=0 <=> a=b (đpcm)

b) a^2+b^2+c^2=ab+bc+ca

<=> 2a^2+2b^2+2c^2=2ab+2bc+2ca

<=> (a^2-2ab+b^2)+(a^2-2ca+c^2)+(b^2-2bc+c^2)=0

<=> (a-b)^2+(a-c)^2+(b-c)^2=0

<=> a-b=0 và a-c=0 và b-c=0

<=> a=b và a=c và b=c

<=> a=b=c (đpcm)