Question

Выражение: 1) (5а+4)^2-40а 2) 16х^2-4(2х+3)^2 3) (3b-2)^2+12b 4) -25x^2+(5х-4)^2 5) (3а-7b)^2+42ab 6) (a+(b-c))^2 7) (2a+6b)^2-24ab 8) (a-(b+c))^2

Answer 1

1) $(5a+4)^{2}-40a= 25a^{2}+40a+16-40a= 25a^{2}+16$
2) $16x^{2}-4*(2x+3)^{2}= 16x^{2}-16x^{2}-48x-36=-48x-36=$$-12(4x+3)$
3) $(3b-2)^{2} +12b=9 b^{2} -12b+4+12b=9 b^{2} +4$
4) $-25 x^{2}+(5x-4)^{2}=-25x^{2}+25 x^{2} -40x+16=16-40x=8(2-5x)$
5) $(3a-7b)^{2}+42ab=9 a^{2} -42ab+49 b^{2}+42ab=9 a^{2} +49 b^{2}$
6) $(a+(b-c))^{2}= (a+b-c)^{2}=(a+b-c)*(a+b-c)=$$a^{2}+ab-ac+ab+ b^{2}-bc-ac-bc+c^{2}=$$a^{2}+b^{2}+c^{2}+2ab-2ac-2bc$
7) $(2a+6b)^{2}-24ab=4 a^{2} +24ab+36 b^{2}-24ab=4a^{2} +36 b^{2}=$$4(a^{2} +9 b^{2})$
8) $(a-(b+c))^{2} = (a-b-c)^{2} =(a-b-c)*(a-b-c)=$$(a-b-c)*(a-b-c)= a^{2}-ab-ac-ab+ b^{2} +bc-ac+bc+ c^{2} =$$a^{2}-ab-ac-ab+ b^{2} +bc-ac+bc+ c^{2} =$$a^{2}+b^{2}+c^{2}-2ab-2ac+2bc$