Ребята! Я в этом ничего не смыслю! Помогите решить, пожалуйста.
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они все решаются однотипно
решу последнее
((3x +y) ^ 2 - (3x - y) ^ 2) / 3xy =
= (9x^2 - 9x^2 + y^2 - y^2 + 6xy + 6xy) / 3xy =
12xy/3xy = 4
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Answers & Comments
1)
2)
3)
они все решаются однотипно
решу последнее
((3x +y) ^ 2 - (3x - y) ^ 2) / 3xy =
= (9x^2 - 9x^2 + y^2 - y^2 + 6xy + 6xy) / 3xy =
12xy/3xy = 4