Объяснение:
4.
log₀,₅(4-x)≥log₀,₅2-log₀,₅(x-1)
ОДЗ: 4-x>0 x<4 x-1>0 x>1 ⇒ x∈(1;4).
log₀,₅(4-x)-log₀,₅2+log₀,₅(x-1)≥0
log₀,₅((4-x)*(x-1)/2)≥0
(4-x)*(x-1)/2≤0,5⁰
(4-x)*(x-1)/2≤1
(4-x)*(x-1)/2-1≤0
((4x-4-x²+x)-2)/2≤0 |×2
4x-4-x²+x-2≤0
-x²+5x-6≤0 |×(-1)
x²-5x+6≥0
x²-5x+6=0 D=1
x₁=2 x₂=3 ⇒
(x-2)(x-3)≥0
-∞__+__2__-__3__+__+∞ ⇒ x∈(-∞;2]U[3;+∞).
Учитывая ОДЗ:
ответ: x∈(1;2]U[3;4).
5.
{xy+x+y=15 {xy+x+y=15
{x²y+xy²=54 {xy*(x+y)=54
Пусть x+y=t, a xy=v ⇒
{t+v=15 {v=15-t {v=15-t {v=15-t
{tv=54 {t*(15-t)=54 {15t-t²-54=0 |×(-1) {t²-15t+54=0
t²-15t+54=0 D=9 √D=3
{t₁=x+y=6 {y=6-x {y=6-x {y=6-x
{v₁=xy=9 {x*(6-x)=9 {6x-x²-9=0 |×(-1) {x²-6x+9=0
{y=6-x {y=6-x y₁=3
{(x-3)²=0 {x-3=0 x₁=3.
{t₂=x+y=9 {y=9-x {y=9-x {y=9-x
v₂=xy=6 {x*(9-x)=6 {9x-x²-6=0 |(×-1) {x²-9x+6=0 D=57
y₂=(9+√57)/2 y₃=(9-√57)/2
x₂=(9-√57)/2 x₃=(9+√57)/2.
ответ: x₁=3 y₁=3 x₂=(9-√57)/2 y₂=(9+√57)/2
x₃=(9+√57)/2 y₃=(9-√57)/2.
6.
y=eˣ*cosx
y'=(eˣ)'*cosx+eˣ*(cosx)'=eˣ*cosx+eˣ*(-sinx)=eˣ*cosx-eˣ*sinx
y'=eˣ*(cosx-sinx).
