Берешь любые значения x,лучше маленькие для удобства, подстпвляешь и находишь у, два числа это координаты , с помощью них можно построить точку на координатной плоскости. получаешь много точек и соединяешь их, вуоля
4)
![x^2-4|x|\ \textless \ 12](https://tex.z-dn.net/?f=x%5E2-4%7Cx%7C%5C+%5Ctextless+%5C+12)
Рассмотрим два случая:
![\left \{ {{x \geq 0} \atop {x^2-4x-12\ \textless \ 0}} \right. \\ \left \{ {{x \geq 0} \atop {(x-6)(x+2)\ \textless \ 0}} \right. \\ \left \{ {{x \geq 0} \atop {x \in (-2; 6)}} \right. \\ x\in[0;6)](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx+%5Cgeq+0%7D+%5Catop+%7Bx%5E2-4x-12%5C+%5Ctextless+%5C+0%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx+%5Cgeq+0%7D+%5Catop+%7B%28x-6%29%28x%2B2%29%5C+%5Ctextless+%5C+0%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx+%5Cgeq+0%7D+%5Catop+%7Bx+%5Cin+%28-2%3B+6%29%7D%7D+%5Cright.++%5C%5C+x%5Cin%5B0%3B6%29)
и
![\left \{ {{x\ \textless \ 0} \atop {x^2+4x-12\ \textless \ 0}} \right. \\ \left \{ {{x\ \textless \ 0} \atop {(x-2)(x+6)\ \textless \ 0}} \right. \\ \left \{ {{x\ \textless \ 0} \atop {x\in(-6; 2)}} \right. \\ x\in(-6; 0)](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+0%7D+%5Catop+%7Bx%5E2%2B4x-12%5C+%5Ctextless+%5C+0%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+0%7D+%5Catop+%7B%28x-2%29%28x%2B6%29%5C+%5Ctextless+%5C+0%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+0%7D+%5Catop+%7Bx%5Cin%28-6%3B+2%29%7D%7D+%5Cright.+%5C%5C+x%5Cin%28-6%3B+0%29+)
Отсюда
x∈(-6; 6)
5)
![x^2-5x+9\ \textgreater \ |x-6|](https://tex.z-dn.net/?f=x%5E2-5x%2B9%5C+%5Ctextgreater+%5C+%7Cx-6%7C)
Опять рассмотрим два случая:
![\left \{ {{x-6 \geq 0} \atop {x^2-5x+9\ \textgreater \ x-6}} \right. \\ \left \{ {{x \geq 6} \atop {x^2-6x+15\ \textgreater \ 0}} \right. \\ \left \{ {{x \geq 6} \atop {x\in(-\infty; +\infty)}} \right. \\ x\in[6; +\infty)](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx-6+%5Cgeq+0%7D+%5Catop+%7Bx%5E2-5x%2B9%5C+%5Ctextgreater+%5C+x-6%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx+%5Cgeq+6%7D+%5Catop+%7Bx%5E2-6x%2B15%5C+%5Ctextgreater+%5C+0%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx+%5Cgeq+6%7D+%5Catop+%7Bx%5Cin%28-%5Cinfty%3B+%2B%5Cinfty%29%7D%7D+%5Cright.++%5C%5C+x%5Cin%5B6%3B+%2B%5Cinfty%29)
и
![\left \{ {{x-6\ \textless \ 0} \atop {x^2-5x+9 \ \textgreater \ 6 - x}} \right. \\ \left \{ {{x\ \textless \ 6} \atop {x^2-4x+3\ \textgreater \ 0}} \right. \\ \left \{ {{x\ \textless \ 6} \atop {(x-3)(x-1)\ \textgreater \ 0}} \right. \\ \left \{ {{x\ \textless \ 6} \atop {x\in(-\infty; 1)\cup(3;+\infty)}} \right. \\ x\in(-\infty;1)\cup(3;6)](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx-6%5C+%5Ctextless+%5C+0%7D+%5Catop+%7Bx%5E2-5x%2B9+%5C+%5Ctextgreater+%5C++6+-+x%7D%7D+%5Cright.+%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+6%7D+%5Catop+%7Bx%5E2-4x%2B3%5C+%5Ctextgreater+%5C+0%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+6%7D+%5Catop+%7B%28x-3%29%28x-1%29%5C+%5Ctextgreater+%5C+0%7D%7D+%5Cright.+%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+6%7D+%5Catop+%7Bx%5Cin%28-%5Cinfty%3B+1%29%5Ccup%283%3B%2B%5Cinfty%29%7D%7D+%5Cright.++%5C%5C+x%5Cin%28-%5Cinfty%3B1%29%5Ccup%283%3B6%29)
Отсюда
x∈(-∞; 1)∪(3; +∞)
6.
![x^2+2|x-1|+7 \leq 4|x-2|](https://tex.z-dn.net/?f=x%5E2%2B2%7Cx-1%7C%2B7+%5Cleq+4%7Cx-2%7C)
Здесь уже рассмотрим 3 случая - x относительно чисел 1 и 2.
![\left \{ {{x\ \textless \ 1} \atop {x^2-2(x-1)+7 \leq -4(x-2)}} \right. \\ \left \{ {{x\ \textless \ 1} \atop {x^2-2x+2+7+4x-8 \leq 0}} \right. \\ \left \{ {{x\ \textless \ 1} \atop {x^2+2x+1 \leq 0}} \right. \\ \left \{ {{x\ \textless \ 1} \atop {x=-1}} \right. \\ x=-1](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+1%7D+%5Catop+%7Bx%5E2-2%28x-1%29%2B7+%5Cleq+-4%28x-2%29%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+1%7D+%5Catop+%7Bx%5E2-2x%2B2%2B7%2B4x-8+%5Cleq+0%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+1%7D+%5Catop+%7Bx%5E2%2B2x%2B1+%5Cleq+0%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%5C+%5Ctextless+%5C+1%7D+%5Catop+%7Bx%3D-1%7D%7D+%5Cright.++%5C%5C+x%3D-1)
Отсюда
x=-1
2√(1 -4^x)/(4^(x-1) - 63*√(4^x/(1-4^x) ≤ 3√63;
4√(1 - 4^x)/4^x -63*√(4^x/(1-4^x) ≤ 3√63;
ОДЗ: 1 - 4^x ≠ 0 ⇒ x≠0. [ 4^x ≠ 1; 4^x ≠ 4^0; x≠0 ].
4t -63/t ≤ 3√63
4t² -3√63 *t -63 ≤ 0; (
4(t +√63/4)(t -√63) ≤ 0;
- √63/4 ≤ t ≤√63;
- √63/4 ≤√ ((1-4^x)/4^x) ≤√63;
0 ≤√ ((1-4^x)/4^x) ≤√63;
0 ≤ (1-4^x)/4^x ≤63;
0 ≤ 1-4^x ≤63*4^x ;
1/64 ≤4^x ≤1;
4^(-3) ≤ 4^x ≤ 4^ 0;
-3 ≤x ≤ 0 , но x =0 ∉ ОДЗ , поэтому ,
-3≤x < 0.
ответ: x∈ [ -3 ;0).
<span>найдите первые пять членов геометрической прогрессии. bn, если b1=0,14, q=V2</span>
Там не все там есть и лишние извини