Question

Cho 3 số a,b,c thỏa mãn a+b-c=0 và \(a^2+b^2+c^2=10\)

Tính \(a^4+b^4+c^4\)

Mong các bạn giúp mình

Answer 1

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

\(\Leftrightarrow a^2+b^2+c^2=-2ab-2bc-2ca=10\) (do a²+b²+c²=10)

\(\Leftrightarrow-2\left(ab+bc+ca\right)=10\Leftrightarrow ab+bc+ca=-5\)

\(\Leftrightarrow\left(ab+bc+ca\right)^2=25\Leftrightarrow a^2b^2+b^2c^2+c^2a^2+2abc\left(a+b+c\right)=25\)

\(\Leftrightarrow a^2b^2+b^2c^2+c^2a^2=25\) (do a+b+c=0)

Lại có: \(a^2+b^2+c^2=10\Leftrightarrow\left(a^2+b^2+c^2\right)^2=100\Leftrightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+c^2a^2\right)=100\)

\(\Leftrightarrow a^4+b^4+c^4+2.25=100\Leftrightarrow a^4+b^4+c^4=50\)