Tg2a=2tga/(1-tg^2a)=2*49/7*48=7/24
6 и 4
6х4=24
6+4=10
вот как - то так
Из уравнения (2) выразим переменную х
![x= \frac{y^2-3}{y}](https://tex.z-dn.net/?f=x%3D+%5Cfrac%7By%5E2-3%7D%7By%7D+)
Подставляем в (1) уравнение
![(\frac{y^2-3}{y})^2+2(-3+y^2)=16|\cdot y^2\\ (y^2-3)^2+2y^2(y^2-3)=16y^2\\ y^4-6y^2+9+2y^4-6y^2=16y^2\\ 3y^4-28y^2+9=0|:3](https://tex.z-dn.net/?f=%28%5Cfrac%7By%5E2-3%7D%7By%7D%29%5E2%2B2%28-3%2By%5E2%29%3D16%7C%5Ccdot+y%5E2%5C%5C+%28y%5E2-3%29%5E2%2B2y%5E2%28y%5E2-3%29%3D16y%5E2%5C%5C+y%5E4-6y%5E2%2B9%2B2y%5E4-6y%5E2%3D16y%5E2%5C%5C+3y%5E4-28y%5E2%2B9%3D0%7C%3A3)
Пусть y^2=t, (t≥0)
3t^2-28t+9=0
t1 = 1/3
t2= 9
Возвращаемся к замене
![y^2= \frac{1}{3} \\ y=\pm \frac{ \sqrt{3} }{3} \\ x=\pm\frac{8 \sqrt{3} }{3}\\\\ y^2=9\\ y=\pm 3\\ x=\pm2](https://tex.z-dn.net/?f=y%5E2%3D+%5Cfrac%7B1%7D%7B3%7D+%5C%5C+y%3D%5Cpm+%5Cfrac%7B+%5Csqrt%7B3%7D+%7D%7B3%7D+%5C%5C+x%3D%5Cpm%5Cfrac%7B8+%5Csqrt%7B3%7D+%7D%7B3%7D%5C%5C%5C%5C+y%5E2%3D9%5C%5C+y%3D%5Cpm+3%5C%5C+x%3D%5Cpm2)
Ответ:
![(\pm2;\pm3),\,(\pm\frac{8 \sqrt{3} }{3};\pm\frac{ \sqrt{3} }{3})](https://tex.z-dn.net/?f=%28%5Cpm2%3B%5Cpm3%29%2C%5C%2C%28%5Cpm%5Cfrac%7B8+%5Csqrt%7B3%7D+%7D%7B3%7D%3B%5Cpm%5Cfrac%7B+%5Csqrt%7B3%7D+%7D%7B3%7D%29)
√32*√98=√32*98=√3136=56
√12*√18*√216=√12*18*√216=√216*√216=216