Предмет: Алгебра, автор: sergejpankratov46

Помогите с заданием, пожалуйста.

Приложения:

Ответы

Автор ответа: kuanramazan07
1

Ответ:

3(2х-1)+7=5(х-1)+7 (3x-8)(7x+5)=(3x-8)^2

21x^2+15x-56x-40=9x^2-48x+64

21x^2-41x-40=9x^2-48x+64

21x^2-9x^2-41x+48x-40-64=0

12x^2+7x-104=0

D=49-4*12*(-104)=49+4992=5041=71^2

x1=(-7+71)/24=64/24=32/12=16/6=8/3

x2=(-7-71)/24=-78/24=-39/12=-13/4|2x-3|=2x+3

2x-3=2x+3

-3=3

нет решений

-2x+3=2x+3

-4x=0

x=0Раскройте скобки

\textcolor{#C58AF9}{(3x-5)(4x+8)}=12x(x-2)+44

\textcolor{#C58AF9}{3x(4x+8)-5(4x+8)}=12x(x-2)+44

2

Раскройте скобки

\textcolor{#C58AF9}{3x(4x+8)}-5(4x+8)=12x(x-2)+44

\textcolor{#C58AF9}{12x^{2}+24x}-5(4x+8)=12x(x-2)+44

3

Раскройте скобки

12x^{2}+24x\textcolor{#C58AF9}{-5(4x+8)}=12x(x-2)+44

12x^{2}+24x\textcolor{#C58AF9}{-20x-40}=12x(x-2)+44

4

Объедините подобные члены

12x^{2}+\textcolor{#C58AF9}{24x}\textcolor{#C58AF9}{-20x}-40=12x(x-2)+44

12x^{2}+\textcolor{#C58AF9}{4x}-40=12x(x-2)+44

5

Раскройте скобки

12x^{2}+4x-40=\textcolor{#C58AF9}{12x(x-2)}+44

12x^{2}+4x-40=\textcolor{#C58AF9}{12x^{2}-24x}+44

Решение

x=3

6х-3+7=5х-5+7

6х+4=5х+2

6х-5х=2-4

х=-2

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