2. y = -3cos6x + 13
область значений (E(y)) обычной функции cosx [-1;1]
т.е. -1 ≤ cosx ≤ 1
но наша функция имеет немного другой вид, значит, нужно всё двойное неравенство умножить на -3
3 ≥ -3cos6x ≥ -3
и прибавим 13
3 + 13 ≥ -3cos6x +13 ≥ -3 + 13
16 ≥ -3cos6x + 13 ≥ 10
10 ≤ -3cosx + 13 ≤ 16
т.е. E (y) = [10;16]
1) <span>6√2+5√18=</span>6<span>√2+5<span>√(2*9)=6√2+5*3√2=6√2+15√2=21√2</span></span>
<span><span>2) <span>5√75-2√27=<span>5√(25*3)-2√(3*9)=25<span>√3-6√3=19<span>√3</span></span></span></span></span></span><span><span><span><span><span><span>3) <span>√2+√50-√18=<span>√2+√(25*2)-√(2*9)=<span>√2+5√2-3√2=3<span>√2</span></span></span></span></span></span></span></span></span></span><span><span><span><span><span><span><span><span><span><span>4) <span>3√20+5√45-2√80=<span>3√(4*5)+5√(9*5)-2√(5*16)=<span>6√5+15√5-8√5=1<span>3√5</span></span></span></span></span></span></span></span></span></span></span></span></span></span><span><span><span><span><span><span><span><span><span><span><span><span><span><span>5) <span>2√48+√27+√12=<span>2√(16*3)+√(9*3)+√(3*4)=<span>8√3+3√3+2√3=13<span>√3</span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span><span><span><span><span><span><span><span><span><span><span><span><span><span><span><span><span><span><span>6) <span>(√3-1) (√3+1)= по формуле разность квадратов=3-1=2</span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span></span>
Б)5x+10=x-2
5x-x=-2-10
4x=-12
x=-12:4
x=-3
-4х+17>2x+5 переносим х в левую сторону, а числа - в правую
Y=log₀,₅(x²-7x+12)
ОДЗ-?
x²-7x+12>0
x₁=3; x₂=4 ( корни найдены по т.Виета)
+ - +
________(3)______________(4)__________
ОДЗ: x∈(-∞;3)U(4;+∞)