y=sin(x/2)+3)
y ' = cos(x/2+3)/2
y^2=sin^(2)(x/2+3)
(2y ')^2=4*cos^(2)(x/2 +3)*(1/4)=cos^(2)(x/2+3)
тогда
<span>y^2 +( 2y' )^2=sin^(2)(x/2+3)+cos^(2)(x/2+3)=1</span>
<span>использовали формулу</span>
sin^2(x)+cos^2(x)=1
<span>1+2log</span>ₓ<span>5= log</span>₅<span>x
1) Сначала ОДЗ
x > 0
х </span>≠ 1
<span>2) теперь решаем:
1 + 2*log</span>₅5/log₅x = log₅x | * log₅x ≠ 0
<span>log</span>₅x +2log₅x = log₅²x
<span>log</span>₅x = t
<span>t +2t - t</span>² = 0
<span>t</span>² -3t = 0
<span>t(t - 3) = 0
a) t = 0 б) t = 3
log</span>₅x = 0 log₅x = 3
<span>не подходит х = 5</span>³ = 125<span>
к ОДЗ
Ответ : 125
</span>
.............................дм
Log₂ log₃ 81 = log₂ log₃ (3⁴) = log₂ 4 log₃ 3 = log₂ (2²) = 2
log₃ log₂ 8 = log₃ log₂ (2³) = log₃ 3 log₂ 2 = 1*1 = 1