Ta có :

\(\left(a+b+c\right)=0\)

\(\Leftrightarrow\left(a+b+c\right)^2=0\)

\(\Leftrightarrow a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Leftrightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)

\(\Leftrightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)

\(\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\left(1\right)\)

\(\Leftrightarrow a^4+b^4+c^4=4\left(a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Leftrightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\left(2\right)\) (vì \(a+b+c=0\))

\(\left(1\right)+\left(2\right)\Rightarrow2\left(a^4+b^4+c^4\right)=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)

\(\Rightarrow\left(a^4+b^4+c^4\right)=2\left(ab+bc+ca\right)^2\)

\(\Rightarrow dpcm\)

Câu a/ Thì chứng minh ở dưới rồi nhé e

b/ Ta cần chứng minh

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

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

\(\Leftrightarrow2abc\left(a+b+c\right)=0\)(đúng)

=> ĐPCM

c/ Ta có

\(\frac{\left(a^2+b^2+c^2\right)^2}{2}=\frac{a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)}{2}=a^4+b^4+c^4\)

Cái này là áp dụng câu a vô nhé e

Bài 3: Cho a + b + c = 0. Chứng minh a^4 + b^4 + c^4 bằng mỗi biểu thức:
a) 2(a^2b^2 + b^2c^2 + c^2a^2)
b) 2( ab + bc + ca)^2
c) (a^2 + b^2 + c^2)^2 / 2

Ta có :

\(a+b+c=0\)

\(\Rightarrow\left(a+b+c\right)^2=0^2\)

\(a^2+b^2+c^2+2\left(ab+bc+ca\right)=0\)

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)

\(\Rightarrow\left(a^2+b^2+c^2\right)^2=\left[-2\left(ab+bc+ca\right)\right]^2\)

\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4\left(a^2b^2+b^2c^2+a^2c^2+2ab^2c+2abc^2+2a^2bc\right)\)

\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4a^2b^2+4b^2c^2+4a^2c^2+8ab^2c+8abc^2+8a^2bc\)

\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4a^2b^2+4b^2c^2+4a^2c^2+8abc\left(a+b+c\right)\)

Mà \(a+b+c=0\)

\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2=4a^2b^2+4b^2c^2+4a^2c^2\)

Bớt cả 2 vế đi\(2a^2b^2+2b^2c^2+2a^2c^2\)có :

\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2a^2c^2\)

Lại cộng cả 2 vế cho \(a^4+b^4+c^4;\)có :

\(\Rightarrow a^4+b^4+c^4+a^4+b^4+c^4=+a^4+b^4+c^4+2a^2b^2+2b^2c^2+2a^2c^2\)

\(\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

\(\Rightarrow a^4+b^4+c^4=\frac{\left(a^2+b^2+c^2\right)^2}{2}\)

Vậy ...

a) Ta có: \(a+b+c=0\)

\(\Rightarrow2abc\left(a+b+c\right)=0\)

\(\Rightarrow2a^2bc+2ab^2c+2abc^2=0\)

Ta lại có:

\(a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)^2\)  (cái này bạn tự chứng minh nha)

\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+4a^2bc+4ab^2c+4abc^2\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)

\(\Rightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\left(đpcm\right)\)

b) Ta có: \(a+b+c=0\)

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

\(\Rightarrow a^2=b^2+c^2+2bc\)

\(\Rightarrow a^2-b^2-c^2=2bc\)

\(\Rightarrow a^4+b^4+c^4-2a^2b^2-2a^2c^2+2b^2c^2=4b^2c^2\)

\(\Rightarrow a^4+b^4+c^4=4b^2c^2+2a^2b^2+2a^2c^2-2b^2c^2\)

\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2\)

\(\Rightarrow a^4+b^4+c^4+a^4+b^4+c^4=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2\)

\(\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

\(\Rightarrow a^4+b^4+c^4=\frac{\left(a^2+b^2+c^2\right)^2}{2}\left(đpcm\right)\)

Chúc bạn học tốt và tíck cho mìk vs nhé!

Cảm ơn bạn 

\(\left(ab+bc+ac\right)^2=a^2b^2+b^2c^2+c^2a^2\\ \Leftrightarrow a^2b^2+b^2c^2+a^2c^2+2\left(ab^2c+abc^2+a^2bc\right)=a^2b^2+b^2c^2+c^2a^2\\ \Leftrightarrow2\left(ab^2c+abc^2+a^2bc\right)=0\\ \Leftrightarrow abc\left(a+b+c\right)=0\left(đpcm;a+b+c=0\right)\)

Ta có : a + b + c = 0

( a + b + c )\(^2\) = 0

\(a^2+b^2+c^2+2ab+2bc+2ca=0\)

Nên : \(a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)

\(\left(a^2+b^2+c^2\right)^2=4\left(ab+bc+ca\right)^2\)

\(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left(ab+bc+ca\right)^2\)

\(a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left(a^2b^2+b^2c^2+c^2a^2+2ab^2c+2abc^2+2a^2bc\right)\)

\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+8ab^2c+8abc^2+8a^2bc\)

\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+8abc\left(b+c+a\right)\)

\(a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2\)

Lại có : \(2\left(ab+bc+ca\right)^2\)

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

\(=2a^2b^2+2b^2c^2+2c^2a^2+4ab^2c+4abc^2+4a^2bc\)

\(=2a^2b^2+2b^2c^2+2c^2a^2+4abc\left(b+c+a\right)\)

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

Vì : \(2a^2b^2+2b^2c^2+2c^2a^2=2a^2b^2+2b^2c^2=2c^2a^2\)

Vậy \(a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)

a) Ta có: \(a+b+c=0\)

\(\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)

\(\Rightarrow a^2+b^2+c^2=-2\left(ab+bc+ca\right)\)

\(\Rightarrow a^4+b^4+c^4+2a^2b^2+2b^2c^2+2c^2a^2=4\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left[a^2b^2+b^2c^2+c^2a^2+2abc\left(b+a+c\right)\right]\)

\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=4\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow a^4+b^4+c^4=4\left(a^2b^2+b^2c^2+c^2a^2\right)-2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)\)

b) Ta có: \(a+b+c=0\)

\(\Rightarrow2abc\left(a+b+c\right)=0\)

\(\Rightarrow2a^2bc+2ab^2c+2abc^2=0\)

Ta lại có:

\(a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2\right)^2\)(chứng minh câu a)

\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2b^2c^2+2c^2a^2+4a^2bc+4ab^2c+4abc^2\)

\(\Rightarrow a^4+b^4+c^4=2\left(a^2b^2+b^2c^2+c^2a^2+2a^2bc+2ab^2c+2abc^2\right)\)

\(\Rightarrow a^4+b^4+c^4=2\left(ab+bc+ca\right)^2\)

c) Ta có: \(a+b+c=0\)

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

\(\Rightarrow a^2=b^2+c^2+2bc\)

\(\Rightarrow a^2-b^2-c^2=2bc\)

\(\Rightarrow a^4+b^4+c^4-2a^2b^2-2a^2c^2+2b^2c^2=4b^2c^2\)

\(\Rightarrow a^4+b^4+c^4=4b^2c^2+2a^2b^2+2a^2c^2-2b^2c^2\)

\(\Rightarrow a^4+b^4+c^4=2a^2b^2+2a^2c^2+2b^2c^2\)

\(\Rightarrow a^4+b^4+c^4+a^4+b^4+c^4=a^4+b^4+c^4+2a^2b^2+2a^2c^2+2b^2c^2\)

\(\Rightarrow2\left(a^4+b^4+c^4\right)=\left(a^2+b^2+c^2\right)^2\)

\(\Rightarrow a^4+b^4+c^4=\left(a^2+b^2+c^2\right):2\)

(Nhớ k cho mình với nhá!)