Пусть y = uv, тогда y' = u'v + uv':
Решим левый интеграл:
cosx = \frac{1-t^2}{1+t^2} => dx = \frac{2}{1+t^2}dt\\ \int \frac{2(1+t^2)}{(1+t^2)(1-t^2)} dt = \int \frac{2}{(1-t)(1+t)}dt = \int ( \frac{1}{1-t} + \frac{1}{1+t})dt = ln(1-t)+ln( 1+t) = ln|1-t^2| = ln|1-tg^2\frac{x}{2}| \\" class="latex-formula" id="TexFormula2" src="https://tex.z-dn.net/?f=%5Cint%20%5Cfrac%7Bdx%7D%7Bcosx%7D%3B%5C%5C%20tg%5Cfrac%7Bx%7D%7B2%7D%3Dt%20%3D%3E%20cosx%20%3D%20%5Cfrac%7B1-t%5E2%7D%7B1%2Bt%5E2%7D%20%3D%3E%20dx%20%3D%20%5Cfrac%7B2%7D%7B1%2Bt%5E2%7Ddt%5C%5C%20%20%5Cint%20%5Cfrac%7B2%281%2Bt%5E2%29%7D%7B%281%2Bt%5E2%29%281-t%5E2%29%7D%20dt%20%3D%20%5Cint%20%5Cfrac%7B2%7D%7B%281-t%29%281%2Bt%29%7Ddt%20%3D%20%5Cint%20%28%20%5Cfrac%7B1%7D%7B1-t%7D%20%2B%20%5Cfrac%7B1%7D%7B1%2Bt%7D%29dt%20%3D%20ln%281-t%29%2Bln%28%201%2Bt%29%20%3D%20ln%7C1-t%5E2%7C%20%3D%20ln%7C1-tg%5E2%5Cfrac%7Bx%7D%7B2%7D%7C%20%20%5C%5C" title="\int \frac{dx}{cosx};\\ tg\frac{x}{2}=t => cosx = \frac{1-t^2}{1+t^2} => dx = \frac{2}{1+t^2}dt\\ \int \frac{2(1+t^2)}{(1+t^2)(1-t^2)} dt = \int \frac{2}{(1-t)(1+t)}dt = \int ( \frac{1}{1-t} + \frac{1}{1+t})dt = ln(1-t)+ln( 1+t) = ln|1-t^2| = ln|1-tg^2\frac{x}{2}| \\">
Возвращаемся к исходному:
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).