BAI 3 :quy đồng lên ta được a^3/abc+b^3/abc+c^3/abc=(a^3+b^3+c^3)/abc
ta có (a+b)^3=a^3+3a^2b+3ab^2+b^3=>a^3+b^3=(a+b)^3-3ab(a+b)
=>a^3+b^3+c^3=(a+b+c)^3+3(a+b)c(a+b+c)=0+0=0
=>A=0/ABC=0
BAI 4:
theo dinh ly py ta go ta co ah^2=ac^2-hc^2
va ah^2 cung bang ab^2-bh^2
=>2ah^2=ac^2-hc^2+ab^2-bh^2=ab^2+ac^2-hb^2-hc^2=ac^2-bh^2-hc^2
=(bh+ch)^2-bh^2-ch^2=bh^2+2.bh.ch+ch^2-bh^2-ch^2
=2.bh.ch=2ah^2
==>ah^2=bhxch
d. DE cat AM tai O
vi tam giac ahm vuong tai h co ho la trung tuyen nen ho=am/2
ma am=de nen oh=de/2
==>tam giac dhe vuong tai h