(5а+6)^2 -81=(5а)²+2*5а*6+6²-81=25а²+60а+36-81=25а²+60а-45
25-(а+7)^2 =25-(а²+2*а*7+7²)=25-(а²+14а+49)=25-а²-14а-49=-а²-14а-24
9m^2-(1+2m)^2=9м²-1-4м-4м²=5м²-4м-1
(5x-3y)^2 -16x^2=25х²-30ху+9у²-16х²=9х²-30ху+9у²
(5c-3d)^2 -9d^2=25с²-30сд+9д²-9д²=25с-30сд
49m^2-(n+8m)^2=49м<span>²-н²-16нм+64м²=113м²-16нм-н²</span>
4tgx+3ctgx=7
tgx=t
4t+3/t-7=0 4t^2-7t+3=0 t=1 t2=3/4
x=П/4+Пn
x=arctg(3/4)+Пn
sin3x+sin7x=-2
sinx>-1;
чтобы сумма синусов была равна -2,необходимо, чтобы
sin3x=-1
sin7x=-1 x=П/2+2Пn
Пусть ![x^2=u, x-1=v](https://tex.z-dn.net/?f=%20x%5E2%3Du%2C%20x-1%3Dv%20)
![log_{2}v+log_{2}(u+\frac{1}{v})\leq 2log_{2}(\frac{u+v}{2})](https://tex.z-dn.net/?f=%20log_%7B2%7Dv%2Blog_%7B2%7D%28u%2B%5Cfrac%7B1%7D%7Bv%7D%29%5Cleq%202log_%7B2%7D%28%5Cfrac%7Bu%2Bv%7D%7B2%7D%29%20)
![log_{2}(uv+1)\leq log_{2}(\frac{(u+v)^2}{4})](https://tex.z-dn.net/?f=%20log_%7B2%7D%28uv%2B1%29%5Cleq%20log_%7B2%7D%28%5Cfrac%7B%28u%2Bv%29%5E2%7D%7B4%7D%29%20%20)
, следовательно в силу монотонности логарифма:
![uv+1\leq \frac{(u+v)^2}{4}](https://tex.z-dn.net/?f=%20uv%2B1%5Cleq%20%5Cfrac%7B%28u%2Bv%29%5E2%7D%7B4%7D%20%20)
![4uv+4\leq (u+v)^2](https://tex.z-dn.net/?f=%204uv%2B4%5Cleq%20%28u%2Bv%29%5E2%20)
![u^2+2uv+v^2-4uv\geq 4](https://tex.z-dn.net/?f=%20u%5E2%2B2uv%2Bv%5E2-4uv%5Cgeq%204%20)
![u^2-2uv+v^2\geq 4](https://tex.z-dn.net/?f=%20u%5E2-2uv%2Bv%5E2%5Cgeq%204%20)
![(u-v)^2\geq 4](https://tex.z-dn.net/?f=%20%28u-v%29%5E2%5Cgeq%204%20)
![(u-v)^2-2^2\geq 0](https://tex.z-dn.net/?f=%20%28u-v%29%5E2-2%5E2%5Cgeq%200%20)
![(u-v+2)(u-v-2)\geq 0](https://tex.z-dn.net/?f=%20%28u-v%2B2%29%28u-v-2%29%5Cgeq%200%20)
Возвращаемся к замене ![x^2=u, x-1=v](https://tex.z-dn.net/?f=%20x%5E2%3Du%2C%20x-1%3Dv%20)
![(x^2-x+3)(x^2-x-1)\geq 0](https://tex.z-dn.net/?f=%20%28x%5E2-x%2B3%29%28x%5E2-x-1%29%5Cgeq%200%20)
![x^2+x-1\geq 0](https://tex.z-dn.net/?f=%20x%5E2%2Bx-1%5Cgeq%200%20)
или ![x\leq\frac{1-\sqrt{5}}{2}](https://tex.z-dn.net/?f=%20x%5Cleq%5Cfrac%7B1-%5Csqrt%7B5%7D%7D%7B2%7D%20%20%20%20)
Ограничения на логарифмы в переменных
:
![v>0; u+\frac{1}{v}>0; u+v>0](https://tex.z-dn.net/?f=%20v%3E0%3B%20u%2B%5Cfrac%7B1%7D%7Bv%7D%3E0%3B%20u%2Bv%3E0%20)
Отсюда отбрасываем решения, получая:
![x\geq \frac{1+\sqrt{5}}{2}](https://tex.z-dn.net/?f=%20x%5Cgeq%20%5Cfrac%7B1%2B%5Csqrt%7B5%7D%7D%7B2%7D%20%20)
Ответ: ![x\geq \frac{1+\sqrt{5}}{2}](https://tex.z-dn.net/?f=%20x%5Cgeq%20%5Cfrac%7B1%2B%5Csqrt%7B5%7D%7D%7B2%7D%20)
(x-4)²+(y+2)²=R² проходит через (1,-4)→3²+2²=9+4=13=R²
(x-4)²+(y+2)²=(√13)²