12 месяца = 290 рублей + 1 платье
-
6 месяцев = 120 рублей + 2 платья
=
6 месяца = 170 рублей-1 платье
Значит 6 месяцев = 6 месяцев:
120р+2п=170р-1п
3п=50р
п=50/3р
Ответ: 50/3р
P.S. проверьте
<em>4(а + 8) - 7(а - 1) = 12</em>
<em>4а + 32 - 7а + 7 = 12</em>
<em>32 + 7 - 12 = 7а - 4а</em>
<em>27 = 3а</em>
<em>а = 27 : 3</em>
<em><u>а = 9</u></em>
<em>Проверка:</em>
<em>4(9 + 8) - 7(9 - 1) = 12</em>
<em>4 · 17 - 7 · 8 = 12</em>
<em>68 - 56 = 12</em>
<em>12 = 12</em>
<em>Ответ: 9.</em>
1)u=2x^3+1
du=6*x^2dx
u=1+2*0^3=1
u=1+2*1^3=3
3 3
=1/6*S u^4du=4^5/30 / =3^5/30-1^5/30=121/15
1 1
S(0,1)(e^3x)*xdx
Sfdg=fg-Sgdf
f=x
dg=e^3xdx
df=dx
g=(e^3x)/3
=1/3(e^3x)*x(0,1)-1/3S(0,1)e^3xdx=e^3/3-1/3S(0,1)e^udu=e^3/3+((-e^u)/9)(0,3)=e^3/3+1/9(1-e^3)=1/9(1+2e^3)
u=3x
du=3dx
(0,1)-это пределы интегрирования от 0 до 1 например.
u=3-cosx
du=sinxdx
u=3-cos0=2
u=3-cospi/6=3-V3/2
новый интеграл от 2 до 3-V3/2(1/u)du=loq(u)/от 2 до 3-V3/2=loq(3-V3/2)-loq2=loq(1/4*(6-V3)) это ответ
Перепишите подынтегральное выражение:<span><span>e<span>x3</span></span>x=x<span>e<span>x3</span></span></span>Используем интегрирование по частям:<span>∫udv=uv−∫vdu</span><span>пусть <span>u<span>(x)</span>=x</span> и пусть <span>dv<span>(x)</span>=<span>e<span>x3</span></span></span> dx.</span><span>Затем <span>du<span>(x)</span>=1</span> dx.</span><span>Чтобы найти <span>v<span>(x)</span></span>:</span>Не могу найти шаги в поиске этот интеграла.Но интеграл<span><span><span><span>e<span>−<span><span>iπ</span>3</span></span></span>Γ<span>(<span>13</span>)</span></span><span>9Γ<span>(<span>43</span>)</span></span></span>γ<span>(<span>13</span>,<span>x3</span><span>e<span>iπ</span></span>)</span></span>Теперь решаем под-интеграл.Интеграл от произведения функции на константу есть эта константа на интеграл от данной функции:<span>∫<span><span><span>e<span>−<span><span>iπ</span>3</span></span></span>Γ<span>(<span>13</span>)</span></span><span>9Γ<span>(<span>43</span>)</span></span></span>γ<span>(<span>13</span>,<span>x3</span><span>e<span>iπ</span></span>)</span>dx=<span><span><span>e<span>−<span><span>iπ</span>3</span></span></span>Γ<span>(<span>13</span>)</span></span><span>9Γ<span>(<span>43</span>)</span></span></span>∫γ<span>(<span>13</span>,<span>x3</span><span>e<span>iπ</span></span>)</span>dx</span>Не могу найти шаги в поиске этот интеграла.Но интеграл<span>∫γ<span>(<span>13</span>,<span>x3</span><span>e<span>iπ</span></span>)</span>dx</span><span>Таким образом, результат будет: <span><span><span><span>e<span>−<span><span>iπ</span>3</span></span></span>Γ<span>(<span>13</span>)</span></span><span>9Γ<span>(<span>43</span>)</span></span></span>∫γ<span>(<span>13</span>,<span>x3</span><span>e<span>iπ</span></span>)</span>dx</span></span>Теперь упростить:<span>−<span><span><span><span>(−1)</span><span>23</span></span>Γ<span>(<span>13</span>)</span></span><span>9Γ<span>(<span>43</span>)</span></span></span><span>(xγ<span>(<span>13</span>,<span>x3</span><span>e<span>iπ</span></span>)</span>−∫γ<span>(<span>13</span>,<span>x3</span><span>e<span>iπ</span></span>)</span>dx)</span></span>Добавляем постоянную интегрирования:<span>−<span><span><span><span>(−1)</span><span>23</span></span>Γ<span>(<span>13</span>)</span></span><span>9Γ<span>(<span>43</span>)</span></span></span><span>(xγ<span>(<span>13</span>,<span>x3</span><span>e<span>iπ</span></span>)</span>−∫γ<span>(<span>13</span>,<span>x3</span><span>e<span>iπ</span></span>)</span>dx)</span>+<span>constant</span></span>Ответ:<span><span>−<span><span><span><span>(−1)</span><span>23</span></span>Γ<span>(<span>13</span>)</span></span><span>9Γ<span>(<span>43</span>)</span></span></span><span>(xγ<span>(<span>13</span>,<span>x3</span><span>e<span>iπ</span></span>)</span>−∫γ<span>(<span>13</span>,<span>x3</span><span>e<span>iπ</span></span>)</span>dx)</span>+<span>constant</span></span></span>
