Ta có VT
\(\left(a+b+c\right)^2+a^2+b^2+c^2\)
\(=a^2+b^2+c^2+2ab+2bc+2ca+a^2+b^2+c^2\)
\(=\left(a^2+2ab+b^2\right)+\left(b^2+2bc+c^2\right)+\left(a^2+2ac+c^2\right)\)
\(=\left(a+b\right)^2+\left(b+c\right)^2+\left(c+a\right)^2=VP\)\(\left(\text{đ}pcm\right)\)
Chung minh rang
:(a+b+c)^2+a^2+b^2+c^2=(a+b)^2+(b+c)^2+(C+a)^2
chung minh rang: (a+b)2/(a-b)2 +(b+c)2/(b-c)2+(a+c)2/(c-a)2 >=2
cho a,b,c la 3 so khac 0 va a+b+c=0 chung minh rang 1/a^2+b^2-c^2+1/b^2+c^2-a^2+1/c^2+a^2-b^2=0
Cho a/(b+c)+b/(c+a)+c/(a+b)=1. chung minh rang
a^2/(b+c)+b^2/(c+a)+c^2/(a+b)=0
cho a+b+c=1 va 1/a+1/b+1/c=0.Chung minh rang : a^2+b^2+c^2=0
Chung minh rang:
neu: (a-b)2 + (b-c)2 + (c-a)2 = (a+b-2c)2 + (b+c-2a)2 + (c+a-2b)2 thi a=b=c
Cho a,b,c>0.Chung minh rang \(\frac{a^2}{b+2c}+\frac{b^2}{c+2a}+\frac{c^2}{a+2b}\ge\frac{a+b+c}{3}\)
cho a^3 b^3 + a^3 c^3 + b^3 c^3 =3a^2 b^2 c^2. chung minh rang (ab+bc)(bc+ac)(bc+ac)=-a^2 b^2 c^2. Giúp mình đi mình tích cho.
Chung Minh Rang: \(a\left(b-c\right)\left(b+c-a\right)^2+c\left(a-b\right)\left(a+b-c\right)^2=b\left(a-c\right)\left(a+c-b\right)^2\)
