Question

Cho a+b+c=0.chứng minh rằng a^4+b^4+c^4=1/2(a^2+b^2+c^2)^2

Answer 1

a + b + c = 0
<=> (a + b + c)² = 0
<=> a² + b² + c² + 2(ab + bc + ca) = 0
<=> a² + b² + c² = -2(ab + bc + ca) (1)

CẦn chứng minh:

2(a^4 + b^4 + c^4) = (a² + b² + c²)²

<=> 2(a^4 + b^4 + c^4) = a^4 + b^4 + c^4 + 2(a²b² + b²c² + c²a²)

<=> a^4 + b^4 + c^4 = 2(a²b² + b²c² + c²a²)

<=> (a² + b² + c²)² = 4(a²b² + b²c² + c²a²) ---(cộng 2 vế cho 2(a²b² + b²c² + c²a²) )

<=> [-2(ab + bc + ca)]² = 4(a²b² + b²c² + c²a²) ----(do (1))

<=> 4.(a²b² + b²c² + c²a²) + 8.(ab²c + bc²a + a²bc) = 4(a²b² + b²c² + c²a²)

<=> 8.(ab²c + bc²a + a²bc) = 0

<=> 8abc.(a + b + c) = 0

<=> 0 = 0 (đúng), Vì a + b + c = 0

=> Đpcm

Answer 2

a + b + c = 0

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

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

\(\Leftrightarrow a^2+b^2+c^2=-2.\left(ab+bc+ca\right)\left(1\right)\)

Cần phải chứng minh

2.(a⁴ + b⁴ + c⁴)=(a²+b²+c²)

\(\Leftrightarrow\) 2.(a⁴ - b⁴+c⁴)=a⁴+b⁴+c⁴+2.(a²b²+b²c²+c²a²)

\(\Leftrightarrow\)a⁴ +b⁴+c⁴=2.(a²b²+b²c²+c²a²)

\(\Leftrightarrow\) (a² + b² +c² ) = 4(a²b²+b²c² +c²a²)

\(\Leftrightarrow\) [ -2.(ab+bc+ca)² ] = 4(a²b²+b²c² +c²a²)

\(\Leftrightarrow\) 4(a²b²+b²c² +c²a²)+8.(ab²c +bc²a+a²bc)=4.(a²b+b²c²+c²+a²

^{\(\Leftrightarrow\)} 8(ab²c+bc²a+a²bc)=0

\(\Leftrightarrow\)8abc.(a+b+c)=0

\(\Leftrightarrow\) 0 =0 (đúng ) Vì a +b +c =0

=> ĐPCM