Ответ:

Квадраттар айырымының формуласын қолданамыз: \(a^2-b^2=(a-b)(a+b)\).

  1. \(\sqrt{\frac{165^2-124^2}{164}}=\sqrt{\frac{(165-124)(165+124)}{164}}=\sqrt{\frac{41\cdot289}{164}}=\sqrt{\frac{41\cdot17^2}{4\cdot41}}=\sqrt{\frac{17^2}{4}}=\frac{17}{2}=8\frac{1}{2}.\)

  2. \(\sqrt{\frac{98}{176^2-112^2}}=\sqrt{\frac{98}{(176-112)(176+112)}}=\sqrt{\frac{98}{64\cdot288}}=\sqrt{\frac{49}{9216}}=\frac{7}{96}.\)

  3. \(\sqrt{\frac{149^2-76^2}{457^2-384^2}}=\sqrt{\frac{(149-76)(149+76)}{(457-384)(457+384)}}=\sqrt{\frac{73\cdot225}{73\cdot841}}=\sqrt{\frac{15^2}{29^2}}=\frac{15}{29}.\)

  4. \(\sqrt{\frac{145,5^2-96,5^2}{193,5^2-31,5^2}}=\sqrt{\frac{(145,5-96,5)(145,5+96,5)}{(193,5-31,5)(193,5+31,5)}}\)

    \(=\sqrt{\frac{49\cdot242}{162\cdot225}}=\sqrt{\frac{7^2\cdot11^2}{15^2\cdot9^2}}=\frac{77}{135}.\)

Жауабы: 1) \(8\frac{1}{2}\); 2) \(\frac{7}{96}\); 3) \(\frac{15}{29}\); 4) \(\frac{77}{135}\).