ОДЗ: х≠1
(x-2)²(x+4)(1-x)≥0
-(x-2)²(x+4)(x-1)≥0
(x-2)²(x+4)(x-1)≤0
x=2 x=-4 x=1
+ - + +
--------- -4 --------------- 1 ------------- 2 --------------
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x∈(-∞; -4) x= -5 + - - | +
x∈(-4; 1) x=0 + + - | -
x∈(1; 2) x=1.5 + + + | +
x∈(2; +∞) x=3 + + + | +
x∈[ -4; 1)U{2}
Ответ: [-4; 1)U{2}.
Если че приложение photomath
Решение
sin^6(a)+cos⁶(a) = (sin²a)³ + (cos²a)³ =
(sin²a + cos²a)*(sin⁴a - sin²acos²a + cos⁴a) =
= [(sin⁴a + 2sin²acos²a + cos⁴a) - 3sin²acos²a] =
(sin²a + cos²a)² - 3sin<span>²acos²a =
= 1 - </span>3sin²acos²a = 1 - (3/4)*(2sinacosa)*(<span>2sinacosa) =
= 1 - (3/4)*(sin</span>²2a) = 1 - [(1 - cos4a)/2] =
= 1 - 3/8 + (3/8)*cos4a = 5/8 + <span> (3/8)*cos4a = (1/8)*(3cos4a + 5)</span>
4(2x-5)=3(x+2)
8x-20=3x+6
5x=26
x=26/5