• 2022-06-19
    设[tex=1.857x1.357]bZ4KhrFbnCaidqbMGQZfww==[/tex]在[tex=6.0x1.357]jCcQXg2Xc0otX1PZx/i2SQ==[/tex]上连续,证明:若[tex=1.857x1.357]bZ4KhrFbnCaidqbMGQZfww==[/tex]为奇函数,则[tex=6.071x2.714]dZ1ScyWj84mTcoQJz8k2XbF+QVNJUUdRpdrYxpoEEt8=[/tex]
  • 举一反三