Question

\(Cho\) \(a+b+c=0\)

Chứng minh 

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

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

Nhanh nhaaaaa

Answer 1

tự làm đi , đồ ăn sẵn

Answer 2

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

\(=a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)\)=a²b²+b²c²+c²a²+2abc.0=a²b²+b²c²+c²a²=VP

Vậy ta có đpcm

Answer 3

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

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

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

Áp dụng phần a ta đã chứng minh được: \(\left(ab+bc+ca\right)^2=a^2b^2+b^2c^2+c^2a^2\)

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

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