1.
(8х-3)(2х+1)=(4х-1)²
16х²+2х-3 -16х²+8х -1=0
10х -4=0
10х=4
х=4:10
х=0,4
5(х-1)-3(х+2)= -5х
5х -5 -3х -6= -5х
2х+5х=6+5
7х=11
х=11/7
х=1 4/7
Пусть одна часть 3х, а другая 2х.
3х+2х=35
5х=35
Х=7
1)3*7=21(м) - одна часть
2)2*7=14(м)
2*(2sin7a*sina / 2 sin7a*cosa) = 2 * tga.
(x-1)²(x-1+5)≥0 (x-1)²(x+4)≥0 <span>x∈<-4,∞)</span>
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Использованы формулы
2 sin α · cos α = sin (2α) - синус двойного аргумента
sin (-α) = -sin α - нечётность функции sin
sin²α + cos²α = 1 - основное тригонометрическое тождество
a² - b² = (a - b)(a + b) - разность квадратов