Question

help me

Answer 1

Từ a+b+c=0

<=>(a+b+c)²=0

<=>a²+b²+c²+2ab+2ac+2bc=0

<=>1+2(ab+ac+bc)=0

<=>2(ab+ac+bc)=-1

<=>ab+ac+bc=\(\dfrac{-1}{2}\)

<=>(ab+ac+bc)²=\(\left(\dfrac{-1}{2}\right)^2\)

<=>a²b²+b²c²+a²c²+2abac+2abbc+2acbc=\(\dfrac{1}{4}\)

<=>a²b²+b²c²+a²c²+2abc(a+b+c)=\(\dfrac{1}{4}\)

<=>a²b²+b²c²+a²c²=\(\dfrac{1}{4}\)(vì a+b+c=0)

Từ a²+b²+c²=1

<=>(a²+b²+c²)²=1

<=>a⁴+b⁴+c⁴+2a²b²+2b²c²+2a²c²=1

<=>a⁴+b⁴+c⁴+2(a²b²+b²c²+a²c²)=1

<=>a⁴+b⁴+c⁴+2\(\cdot\dfrac{1}{4}\)=1

<=>a⁴+b⁴+c⁴+\(\dfrac{1}{2}=1\)

<=>a⁴+b⁴+c⁴=\(\dfrac{1}{2}\)

Vậy a⁴+b⁴+c⁴=\(\dfrac{1}{2}\)

Answer 2

0,5