<span>x^2-3x+</span>√(<span>3-x)=</span>√(<span>3-x)+10 ОДЗ: 3-x>0 => x<3
x^2-3x+</span>√(3-x)-√(3-x)-10=0
x^2-3x-10=0
x₁+x₂=3
x₁*x₂=-10
x₁=5 - лишний корень, не входит в ОДЗ
x₂=-2
Ответ: х=-2
![\frac{- x^{2} }{2+x} \geq \frac{1}{2} \\ \frac{- x^{2} }{2+x} - \frac{1}{2} \geq 0 \\ \frac{- x^{2}*2 }{2(2+x)} - \frac{2+x}{2(2+x)} \geq 0 \\ \frac{- 2x^{2} -2-x}{2(2+x)} \geq 0 \\ \left \{ {{- 2x^{2} -x-2 \geq 0} \atop {2+x \neq 0}} \right. \\ \left \{ {{2x^{2} +x+2 \leq 0} \atop {x \neq -2}} \right. \\ \\ 2x^{2} +x+2 \leq 0 \\ D=b^{2}-4ac \\ D=1^{2}-4*2*2=1-16=-15 \leq 0](https://tex.z-dn.net/?f=+%5Cfrac%7B-+x%5E%7B2%7D+%7D%7B2%2Bx%7D++%5Cgeq++%5Cfrac%7B1%7D%7B2%7D++%5C%5C++%5Cfrac%7B-+x%5E%7B2%7D+%7D%7B2%2Bx%7D++-+%5Cfrac%7B1%7D%7B2%7D+++%5Cgeq+0+%5C%5C+%5Cfrac%7B-+x%5E%7B2%7D%2A2+%7D%7B2%282%2Bx%29%7D++-+%5Cfrac%7B2%2Bx%7D%7B2%282%2Bx%29%7D+++%5Cgeq+0++%5C%5C+%5Cfrac%7B-+2x%5E%7B2%7D+-2-x%7D%7B2%282%2Bx%29%7D++++%5Cgeq+0+%5C%5C++++%5Cleft+%5C%7B+%7B%7B-+2x%5E%7B2%7D+-x-2+%5Cgeq+0%7D+%5Catop+%7B2%2Bx+%5Cneq+0%7D%7D+%5Cright.+%5C%5C++++%5Cleft+%5C%7B+%7B%7B2x%5E%7B2%7D+%2Bx%2B2++%5Cleq+0%7D+%5Catop+%7Bx+%5Cneq+-2%7D%7D+%5Cright.++%5C%5C++%5C%5C+2x%5E%7B2%7D+%2Bx%2B2++%5Cleq+0+%5C%5C+D%3Db%5E%7B2%7D-4ac++%5C%5C+D%3D1%5E%7B2%7D-4%2A2%2A2%3D1-16%3D-15+%5Cleq+0)
при D<0 вещественных корней нет.
2-2cos^2x+sqrt(2)cosx=0
cosx=t
2-2t^2+sqrt(2)t=0
2t^2-sqrt(2)t-2=0
D=2+16=18
t=[sqrt(2)+-3sqrt(2)]/4
t1=4sqrt(2)/4=sqrt(2)>1 не подходит т.к |cosx|<=1G
t2=-sqrt(2)/2
x=+-3П/4+2Пk