23. Ответ:
24. Ответ:
25.
Ответ:
26.
27.
28.
29.
30.
31.
Номера 32 и 33 уже разложены на множители.
Изначально в 32 номере было выражение: ,
а в номере 33 выражение:
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Answers & Comments
23. Ответ:
24. Ответ:
25.![[(10-(10-n)]*[(10+(10-n)]=n(20-n) [(10-(10-n)]*[(10+(10-n)]=n(20-n)](https://tex.z-dn.net/?f=%5B%2810-%2810-n%29%5D%2A%5B%2810%2B%2810-n%29%5D%3Dn%2820-n%29)
Ответ:
26.
Ответ:
27.![[10(6a+3b)]^2-[9(3a+2b)]^2=(60a+30b-27a-18b)(60a+30b+27a+18b)= [10(6a+3b)]^2-[9(3a+2b)]^2=(60a+30b-27a-18b)(60a+30b+27a+18b)=](https://tex.z-dn.net/?f=%5B10%286a%2B3b%29%5D%5E2-%5B9%283a%2B2b%29%5D%5E2%3D%2860a%2B30b-27a-18b%29%2860a%2B30b%2B27a%2B18b%29%3D)
Ответ:
28.
Ответ:
29.
Ответ:
30.
Ответ:
31.
Ответ:
Номера 32 и 33 уже разложены на множители.
Изначально в 32 номере было выражение:
,
а в номере 33 выражение: