Задание 360
Преобразуйте выражение в многочлен стандартного вида:
- \({\largeа)}\ (m+n)^2+(m-n)^2;\)
- \({\largeб)}\ 2(a-1)^2+3(a-2)^2;\)
- \({\largeв)}\ 5(x-y)^2+(x-2y)^2;\)
- \({\largeг)}\ 4(m-2n)^2-3(3m+n)^2;\)
- \({\largeд)}\ 3(2a-b)^2-5(a-2b)^2;\)
- \({\largeе)}\ 4(3x+4y)^2-7(2x-3y)^2;\)
- \({\largeж)}\ 2(p-3q)^2-4(2p-q)^2-(2q-3p)(p+q);\)
- \({\largeз)}\ 5(n-5m)^2-6(2n-3m)^2-(3m-n)(7m-n);\)
- \({\largeи)}\ (2p-q)^2-2(2p-q)(p-q)+(p-q)^2.\)
Источник заимствования: Алгебра. 7 класс. Учебник для общеобразовательных организаций / С.М. Никольский, М.К. Потапов, Н.Н. Решетников, А.В. Шевкин – Просвещение, 2013. – 104 c. ISBN 978-5-09-027739-6
А+АА-
Решение:
- \({\largeа)}\ (m+n)^2+(m-n)^2=\) \(m^2\) \(+\) \(2mn\) \(+\) \(n^2\) \(+\) \(m^2\) \(-\) \(2mn\) \(+\) \(n^2\) \(=2m^2+2n^2\)
- \({\largeб)}\ 2(a-1)^2+3(a-2)^2=2(a^2-2a+1)+3(a^2-4a+4)=\) \(2a^2\) \(-\) \(4a\) \(+\) \(2\) \(+\) \(3a^2\) \(-\) \(12a\) \(+\) \(12\) \(=5a^2-16a+14\)
- \({\largeв)}\ 5(x-y)^2+(x-2y)^2=5(x^2-2xy+y^2)+x^2-4xy+4y^2=\) \(5x^2\) \(-\) \(10xy\) \(+\) \(5y^2\) \(+\) \(x^2\) \(-\) \(4xy\) \(+\) \(4y^2\) \(=6x^2-14xy+9y^2\)
- \({\largeг)}\ 4(m-2n)^2-3(3m+n)^2=4(m^2-4mn+4n^2)-3(9m^2+6mn+n^2)=\) \(4m^2\) \(-\) \(16mn\) \(+\) \(16n^2\) \(-\) \(27m^2\) \(-\) \(18mn\) \(-\) \(3n^2\) \(={-}23m^2-34mn+13n^2\)
- \({\largeд)}\ 3(2a-b)^2-5(a-2b)^2=3(4a^2-4ab+b^2)-5(a^2-4ab+4b^2)=\) \(12a^2\) \(-\) \(12ab\) \(+\) \(3b^2\) \(-\) \(5a^2\) \(+\) \(20ab\) \(-\) \(20b^2\) \(=7a^2+8ab-17b^2\)
- \({\largeе)}\ 4(3x+4y)^2-7(2x-3y)^2=4(9x^2+24xy+16y^2)-7(4x^2-12xy+9y^2)=\) \(36x^2\) \(+\) \(96xy\) \(+\) \(64y^2\) \(-\) \(28x^2\) \(+\) \(84xy\) \(-\) \(63y^2\) \(=8x^2+180xy+y^2\)
- \({\largeж)}\ 2(p-3q)^2-4(2p-q)^2-(2q-3p)(p+q)=2(p^2-6pq+9q^2)-4(4p^2-4pq+q^2)-(2pq-3p^2+2q^2-3pq)=\) \(2p^2\) \(-\) \(12pq\) \(+\) \(18q^2\) \(-\) \(16p^2\) \(+\) \(16pq\) \(-\) \(4q^2\) \(-\) \(2pq\) \(+\) \(3p^2\) \(-\) \(2q^2\) \(+\) \(3pq\) \(={-}11p^2+5pq+12q^2\)
- \({\largeз)}\ 5(n-5m)^2-6(2n-3m)^2-(3m-n)(7m-n)=5(n^2-10mn+25m^2)-6(4n^2-12mn+9m^2)-(21m^2-7mn-3mn+n^2)=\) \(5n^2\) \(-\) \(50mn\) \(+\) \(125m^2\) \(-\) \(24n^2\) \(+\) \(72mn\) \(-\) \(54m^2\) \(-\) \(21m^2\) \(+\) \(7mn\) \(+\) \(3mn\) \(-\) \(n^2\) \(=50m^2+32mn-20n^2\)
- \({\largeи)}\ (2p-q)^2-2(2p-q)(p-q)+(p-q)^2=4p^2-4pq+q^2-2(2p^2-pq-2pq+q^2)+p^2-2pq+q^2=4p^2-4pq+q^2-2(2p^2-3pq+q^2)+p^2-2pq+q^2=\) \(4p^2\) \(-\) \(4pq\) \(+\) \(q^2\) \(-\) \(4p^2\) \(+\) \(6pq\) \(-\) \(2q^2\) \(+\) \(p^2\) \(-\) \(2pq\) \(+\) \(q^2\) \(=p^2\)
Преобразуем выражение в многочлен стандартного вида, используя формулу квадрата разности:
- \(\phantom{\largeи)}\ (2p-q)^2-2(2p-q)(p-q)+(p-q)^2=((2p-q)-(p-q))^2=(2p-q-p+q)^2=p^2\)
