Предмет: Алгебра, автор: Аноним

Упростите 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

Ответы

Автор ответа: Олюшка38

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) $(a-(b+c))^{2} = (a-b-c)^{2} =(a-b-c)*(a-b-c)=$ $(a-b-c)*(a-b-c)= a^{2}-ab-ac-ab+ b^{2} +bc-ac+bc+ c^{2} =$ $a^{2}-ab-ac-ab+ b^{2} +bc-ac+bc+ c^{2} =$ $a^{2}+b^{2}+c^{2}-2ab-2ac+2bc$

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