Question

Решите пожалуйста 3 и 4 задание по алгебре.

Answer 1

№ 3
4а²(х+7)+3(х+7)=(х+7)(4а²+3)
если х= 1,05 и а= –0,5,  то
(1,05+7)(4·(–0,5)²+3)=8,05·(4·0,25+3)=8,05·(1+3)=8,05·4= 32,2
Ответ: 32,2

№ 4

а) = 5а²b(3+2abc)

б) = (m+n)(3–(m+n))=(m+n)(3–m–n)

в)  = –6x(1–x)–(1–x)² = (1–x)(–6x–(1–x)) = (1–x)(–6x–1+x) = (1–x)(–5x–1)

г)  = 2ay–b+by–2a= (2ay–2a)+(by–b) = 2a(y–1)+b(y–1) = (y–1)(2a+b)