<span>а) (а+4)х²=а²-а-20
</span> (а+4)х²=<span>(а+4)*(a-5)
x</span>²=(a-5) ⇒(a-5)>0 ⇒ a>5
x=√(a-5)
<span>б) (а+1)х²+2(а-1)х+(а+3)=0
</span><span>(а+1)х²+(2а-2)х+(а+3)=0
</span>
D=(2а-2)²-4*(a+1)*(a+3)=4a²-8a+4-(4a²+4a+12a+12)=
= 4a²-8a+4-(4a²+16a+12)= 4a²-8a+4- 4a²-16a -12= -24a-8 = -8(3а+1)
уравнение имеет решение если -8(3a+1)≥0 (3a+1)≤0 а≤ -1/3
x₁=(-(2а-2)-24a-8)/ 2(a+1)=(-26a-6)/2(a+1) =-2(13а+3)/2(a+1)= - (13а-3)/(а+1)
х₂=(-(2а-2)+24a+8)/ 2(a+1)=(22а+10)/2(a+1)=2(11а+5)/2(a+1)=(11а+5)/(а+1)
Х -колличество 2 р. монет
х-12 - колличество 5р. монет составим уровнение
(2*х)+(х-12)*5=178
2х+5х-60=178
7х=178+60
7х=238
х=238/7
х=34
34шт. 2р. монет
34-12=22шт. 5р. монет
Давно подобное не решал. Надеюсь, правильно.
![\dfrac{2}{\log_{2}x}+\dfrac{5}{\log^{2}_{2}x-\log_{2}x^{3}}\leq \dfrac{\log_{2}x}{\log_{2}\left(\dfrac{x}{8}\right)}](https://tex.z-dn.net/?f=%5Cdfrac%7B2%7D%7B%5Clog_%7B2%7Dx%7D%2B%5Cdfrac%7B5%7D%7B%5Clog%5E%7B2%7D_%7B2%7Dx-%5Clog_%7B2%7Dx%5E%7B3%7D%7D%5Cleq+%5Cdfrac%7B%5Clog_%7B2%7Dx%7D%7B%5Clog_%7B2%7D%5Cleft%28%5Cdfrac%7Bx%7D%7B8%7D%5Cright%29%7D)
<u><em>Заметим, что можно выполнить некие равносильные преобразования</em></u>
![\log_{2}x^{3}\Leftrightarrow 3\log_{2}x;~\log_{2}\left(\dfrac{x}{8}\right)\Leftrightarrow\log_{2}x-\log_{2}8=\log_{2}x-3](https://tex.z-dn.net/?f=%5Clog_%7B2%7Dx%5E%7B3%7D%5CLeftrightarrow+3%5Clog_%7B2%7Dx%3B~%5Clog_%7B2%7D%5Cleft%28%5Cdfrac%7Bx%7D%7B8%7D%5Cright%29%5CLeftrightarrow%5Clog_%7B2%7Dx-%5Clog_%7B2%7D8%3D%5Clog_%7B2%7Dx-3)
Тогда, пусть ![\log_{2}x=t](https://tex.z-dn.net/?f=%5Clog_%7B2%7Dx%3Dt)
![\dfrac{2}{t}+\dfrac{5}{t^2-3t}\leq \dfrac{t}{t-3}\bigskip\\\dfrac{2(t-3)+5-t^2}{t(t-3)}\leq 0~|:(-1)\bigskip\\\dfrac{t^2-2t+6-t}{t(t-3)}\geq 0\bigskip\\\dfrac{\left(t-1\right)^2}{t(t-3)}\geq 0\Leftrightarrow t\in\left(-\infty;0\right)\cup\left\{1\right\}\cup\left(3;+\infty\right)](https://tex.z-dn.net/?f=%5Cdfrac%7B2%7D%7Bt%7D%2B%5Cdfrac%7B5%7D%7Bt%5E2-3t%7D%5Cleq+%5Cdfrac%7Bt%7D%7Bt-3%7D%5Cbigskip%5C%5C%5Cdfrac%7B2%28t-3%29%2B5-t%5E2%7D%7Bt%28t-3%29%7D%5Cleq+0~%7C%3A%28-1%29%5Cbigskip%5C%5C%5Cdfrac%7Bt%5E2-2t%2B6-t%7D%7Bt%28t-3%29%7D%5Cgeq+0%5Cbigskip%5C%5C%5Cdfrac%7B%5Cleft%28t-1%5Cright%29%5E2%7D%7Bt%28t-3%29%7D%5Cgeq+0%5CLeftrightarrow+t%5Cin%5Cleft%28-%5Cinfty%3B0%5Cright%29%5Ccup%5Cleft%5C%7B1%5Cright%5C%7D%5Ccup%5Cleft%283%3B%2B%5Cinfty%5Cright%29)
<u><em>Перейдём обратно к иксу</em></u>
![\log_{2}x\in\left(-\infty;0\right)\cup\left\{1\right\}\cup\left(3;+\infty\right)\Leftrightarrow x\in\left(0;1\right)\cup\{2\}\cup\left(8;+\infty\right)](https://tex.z-dn.net/?f=%5Clog_%7B2%7Dx%5Cin%5Cleft%28-%5Cinfty%3B0%5Cright%29%5Ccup%5Cleft%5C%7B1%5Cright%5C%7D%5Ccup%5Cleft%283%3B%2B%5Cinfty%5Cright%29%5CLeftrightarrow+x%5Cin%5Cleft%280%3B1%5Cright%29%5Ccup%5C%7B2%5C%7D%5Ccup%5Cleft%288%3B%2B%5Cinfty%5Cright%29)
<u><em>Ответ.</em></u> ![x\in\left(0;1\right)\cup\{2\}\cup\left(8;+\infty\right)](https://tex.z-dn.net/?f=x%5Cin%5Cleft%280%3B1%5Cright%29%5Ccup%5C%7B2%5C%7D%5Ccup%5Cleft%288%3B%2B%5Cinfty%5Cright%29)