Предмет: Алгебра, автор: lomakovd932

(2x-5)^2-(2x+3)*(2x-3)=0 решить уравнение

Ответы

Автор ответа: b14071669

Ответ:

(2x-5)^2-(2x-3)(2x+3)=0

4х^2-20х+25-4х^2+6х-6х+9=0

4х^2-4х^2-20х+6х-6х+25+9=0

-20х+34=0

-20х=-34

х=-34:(-20)

х=1,7

Ответ:1,7

Объяснение:

Автор ответа: marimalova06

2
(2х-5) - (2х+3)*(2х-3)=0
2 2
4х - 20х+25-(4х -9)=0
2 2
4х - 20х+25-4х +9=0
-20х+34=0
Х=1,7 или дробью 17
_
10

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