dùng thước đo và so sánh BH và HC nếu ab = ac thì có thể suy ra HB = HC không
\(a+b+c=0\Rightarrow\left(a+b+c\right)^2=0\Rightarrow a^2+b^2+c^2+2ab+2ac+2bc=0\)
\(\Rightarrow2ac+2bc+2ab=-14\)
\(\Rightarrow ac+ab+bc=-7\)
\(\left(ac+bc+ab\right)^2=49\)
\(a^2c^2+b^2c^2+a^2b^2+2abc^2+2ab^2c+2a^2bc=49\)
\(\Rightarrow a^2c^2+b^2c^2+a^2b^2+2abc\left(a+b+c\right)=49\)
\(\Rightarrow a^2c^2+b^2c^2+a^2b^2=49\)
Có \(a^2+b^2+c^2=14\Rightarrow\left(a^2+b^2+c^2\right)^2=196\)
\(\Rightarrow a^4+b^4+c^4+2\left(a^2b^2+b^2c^2+a^2c^2\right)=196\)
\(\Rightarrow a^4+b^4+c^4=196-2.49=196-98=98\)
Có a + b + c = 0
=> a = -b - c
=>a2 = (-b - c)2 = (b - c)2 = b2+ 2bc + c2
=>a2 - (b2 + c2)= 2bc
=>[a2-(b2+c2)]2=(2bc)2
=>a4-2a2(b2+c2)+(b2+c2)=4b2c2
=>a4-2a2b2-2a2c2+b4+2b2c2+c4=4b2c2
=>a4+b4+c4=4b2c2+2a2b2+2a2c2-2b2c2
=>a4+b4+c4=2a2b2+2a2c2+2b2c2
=>2(a4+b4+c4)=2a2b2+2a2c2+2b2c2+a4+b4+c4
=>2(a4+b4+c4)=
=>2(a4+b4+c4)=a4+2a2(b2+c2)+(b2+c2)
=>2(a4+b4+c4)=(a2+b2+c2)2
=>2(a4+b4+c4)=2.14=28 (vì a2+b2+c2=14)
=>a4+b4+c4=28 : 2= 14
Vậy m=14
