Решение во вложении......
Решие
<span>sin α/2 =2/√7 , 0<α<п
sin</span>²(α/2) = (1 - cosα)/2
(2/√7)² = (1 - cosα)/2
1 - cosα = 8/7
cosα = 1 - 8/7
cosα = - 1/7
sinα = √(1 - cos²α) = √(1 - (- 1/7)²) = √(1 - 1/49) = √(48/49) =
<span>= 4√3 / 7</span>
2^( 4х - 1 ) + 2^( 4х - 2 ) - 2^( 4х - 3 ) = 160
2^4х•( ( 2^-1 ) + ( 2^( - 2 ) - 2^( - 3 ) ) = 160
2^4х•( ( 1/2 ) + ( 1/4 ) - ( 1/8 )) = 160
1/2 + 1/4 - 1/8 = 4/8 + 2/8 - 1/8 = 5/8
2^4х • ( 5/8 ) = 160
2^4х = 160 : ( 5/8 )
2^4х = 256
2^4х = 2^8
4х = 8
Х = 2
А) x^2/5y и z-3/y^2 <=> (y*x^2) / 5y^2 и (5z-15) / 5y^2