Совокупность систем:
Решим 2 уравнение 1-ой системы
1-4x²≥0
x²≤1/4
|x|≤1/2
x≤1/2 или x≥-1/2
x∈[-1/2;1/2]
Решим 2 уравнение 2-ой системы:
1-4x²≥(3x+1)²
1-4x²≥9x²+6x+1
13x²+6x≤0
x(13x+6)≤0
Решим методом интервалов.
Нули функции:
x=0; x=-6/13
+ - +
-----------o----------------o----------->
-6/13 0
x∈[-6/13;0]
Совокупность:
![\left \{ {{x\in(-\infty;-\frac{1}{3})} \atop {x\in[-\frac{1}{2};\frac{1}{2}\}} \right.\\ \left \{ {{x\in[-\frac{1}{3};+\infty)} \atop {x\in[-\frac{6}{13};0]}} \right.](https://tex.z-dn.net/?f=%5Cleft+%5C%7B+%7B%7Bx%5Cin%28-%5Cinfty%3B-%5Cfrac%7B1%7D%7B3%7D%29%7D+%5Catop+%7Bx%5Cin%5B-%5Cfrac%7B1%7D%7B2%7D%3B%5Cfrac%7B1%7D%7B2%7D%5C%7D%7D+%5Cright.%5C%5C+%5Cleft+%5C%7B+%7B%7Bx%5Cin%5B-%5Cfrac%7B1%7D%7B3%7D%3B%2B%5Cinfty%29%7D+%5Catop+%7Bx%5Cin%5B-%5Cfrac%7B6%7D%7B13%7D%3B0%5D%7D%7D+%5Cright. "\left \{ {{x\in(-\infty;-\frac{1}{3})} \atop {x\in[-\frac{1}{2};\frac{1}{2}\}} \right.\\ \left \{ {{x\in[-\frac{1}{3};+\infty)} \atop {x\in[-\frac{6}{13};0]}} \right.")
Совокупность:
![x\in[-\frac{1}{2};-\frac{1}{3})\\x\in[-\frac{1}{3};0]](https://tex.z-dn.net/?f=x%5Cin%5B-%5Cfrac%7B1%7D%7B2%7D%3B-%5Cfrac%7B1%7D%7B3%7D%29%5C%5Cx%5Cin%5B-%5Cfrac%7B1%7D%7B3%7D%3B0%5D "x\in[-\frac{1}{2};-\frac{1}{3})\\x\in[-\frac{1}{3};0]")
Ответ: x∈[-1/2;0]