(х-у) (х+у) +(х+у) =(х+у) (х-у+1)
![x^4 - 2x^2 -8=0\\ x^2=t\\ t^2-2t-8=0\\ D=(-2)^2-4*1*(-8)=4+32=36\\ t_1=\frac{2+6}{2*1}=\frac{8}{2}=4\\ t_2=\frac{2-6}{2}=-\frac{4}{2}=-2\\ \\ 1) \ x^2=4\\ x=б2\\ \\ 2) \ x^2=-2\\](https://tex.z-dn.net/?f=x%5E4+-+2x%5E2+-8%3D0%5C%5C+x%5E2%3Dt%5C%5C+t%5E2-2t-8%3D0%5C%5C+D%3D%28-2%29%5E2-4%2A1%2A%28-8%29%3D4%2B32%3D36%5C%5C+t_1%3D%5Cfrac%7B2%2B6%7D%7B2%2A1%7D%3D%5Cfrac%7B8%7D%7B2%7D%3D4%5C%5C+t_2%3D%5Cfrac%7B2-6%7D%7B2%7D%3D-%5Cfrac%7B4%7D%7B2%7D%3D-2%5C%5C+%5C%5C+1%29+%5C+x%5E2%3D4%5C%5C+x%3D%D0%B12%5C%5C+%5C%5C+2%29+%5C+x%5E2%3D-2%5C%5C+)
нет корней поскольку
всегда ![\geq 0](https://tex.z-dn.net/?f=%5Cgeq+0)
ответ: ![б 2](https://tex.z-dn.net/?f=%D0%B1+2)
Итак, x1=1+2V3; x2=1-2V3 (V- знак корня)
(x-1-2V3)( x-1+2V3)=x^2-x+(2V3)x-x+1-2V3-(2V3)x+2V3-12=x^2-2x-11
Б) 3(3х-5)/2 - 4х-1/3 +2=0;
6x-15/6-8x-2/6+2=0;
6x-15-8x+2/6 +2;
-2x-13/6 +2;
6(-2x-13)/6 +2*6;
-2x-13+12;
-2x=1
x=-1/2