Question

a) Cho a+b+c=0. chứng minh: a⁴+b⁴+c⁴=2(a²b²+b²c²+a²c²)

b) Nếu (a²+b²+c²)(x²+y²+z²)=(ax+by+cz)² thì a/x=b/y=c/z

Helps, thanks

Answer 1

a) Có a +b +c=0

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

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

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

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

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

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

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

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