Найдите разность многочленов:
- \({\largeа)}\ (2a+5)-(3a-6)=\ \)
- \({\largeб)}\ (5x-y)-(2y-x)=\ \)
- \({\largeв)}\ (a-2b-7)-(a-b-1)=\ \)
- \({\largeг)}\ (2x+y-3)-(y-3+x)=\ \)
- \({\largeд)}\ (3a+4b+5c)-(a+4b+3c)=\ \)
- \({\largeе)}\ (a-3b+2c)-({-}2a-3b-c)=\ \)
Задание 140
Источник заимствования: Алгебра. Рабочая тетрадь. 7 класс. В 2 частях. Часть 1 / М.К. Потапов, А.В. Шевкин – Просвещение, 2018. – 51 c. ISBN 978-5-09-051661-7
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Решение:
- \({\largeа)}\ (2a+5)-(3a-6)=\)\(2a\)\(+\)\(5\)\(-\)\(3a\)\(+\)\(6\)\(={-}a+11\)
- \({\largeб)}\ (5x-y)-(2y-x)=\)\(5x\)\(-\)\(y\)\(-\)\(2y\)\(+\)\(x\)\(=6x-3y\)
- \({\largeв)}\ (a-2b-7)-(a-b-1)=\)\(a\)\(-\)\(2b\)\(-\)\(7\)\(-\)\(a\)\(+\)\(b\)\(+\)\(1\)\(={-}b-6\)
- \({\largeг)}\ (2x+y-3)-(y-3+x)=\)\(2x\)\(+\)\(y\)\(-\)\(3\)\(-\)\(y\)\(+\)\(3\)\(-\)\(x\)\(=x\)
- \({\largeд)}\ (3a+4b+5c)-(a+4b+3c)=\)\(3a\)\(+\)\(4b\)\(+\)\(5c\)\(-\)\(a\)\(-\)\(4b\)\(-\)\(3c\)\(=2a+2c\)
- \({\largeе)}\ (a-3b+2c)-({-}2a-3b-c)=\)\(a\)\(-\)\(3b\)\(+\)\(2c\)\(+\)\(2a\)\(+\)\(3b\)\(+\)\(c\)\(=3a+3c\)