Verified answer

Решение и ответ:

[tex]\displaystyle 1)\;\;{a^{100}}\div{a^{89}}\cdot{a^2}={a^{100-89}}\cdot{a^2}={a^{11}}\cdot{a^2}={a^{11+2}}={a^{13}}[/tex]

[tex]\displaystyle 2)\;\;{b^{98}} \div {b^{88}} \cdot {b^{15}}={b^{98-88}} \cdot {b^{15}}= {b^{10}} \cdot {b^{15}}={b^{10+15}}={b^{25}}[/tex]