Question

这题怎么解呀

Answer 1

解：原式饥带=（7/3-7/2+49/45)x(-6/7)

=7/3x(-6/7)-7/2x(-6/7)+49/45x(-6/7)
=-2+3-7/15x2
=1-7/15x2
=1-14/15
=1/烂汪芦15
答：答案1/陵弊15

Answer 2

证明：作OH⊥BC于H,OG⊥AB于G,则∠OGE=∠销搜OHD=90°,∵∠AOC=180°-（∠OAC+∠OCA）=180°-1/2（∠BAC+∠ACB）=90°+1/2∠B=120°∴∠DOC=60°,∴∠ODB=∠吵基DOC+∠OCD=60°+∠OCD,∠OEA=∠B+∠OCD=60°+∠OCD,∴∠ODB=∠OEA,∵角升斗谨平线AD,CE相交于点O,∴OG=OH,∴△ODH≌△OEG,∴OE=OD.