х1=2;х2=-3
Объяснение:
Для начала представим уравнение х^2+2х+1 в виде квадрата
Будет (х+1)^2
х(х+1)^2=6(х+1)
х(х+1)=6
х^2+х-6=0
D=1+6*4=25
х1=(-1+5)/2=2
х2=(-1-5)/2=-3
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Для решения уравнения будем использовать свойства степеней:
![\displaystyle \boxed{\bf{\frac{1}{a^m}=a^{-m}}}\\ \boxed{\bf{a^m \cdot a^n = a^{m+n}}}](https://tex.z-dn.net/?f=++%5Cdisplaystyle++%5Cboxed%7B%5Cbf%7B%5Cfrac%7B1%7D%7Ba%5Em%7D%3Da%5E%7B-m%7D%7D%7D%5C%5C+%5Cboxed%7B%5Cbf%7Ba%5Em+%5Ccdot+a%5En+%3D+a%5E%7Bm%2Bn%7D%7D%7D+)
____________________________
![\displaystyle 5^{4-3x} \cdot 5^{8x-2}= \frac{1}{125}](https://tex.z-dn.net/?f=+%5Cdisplaystyle+5%5E%7B4-3x%7D+%5Ccdot+5%5E%7B8x-2%7D%3D+%5Cfrac%7B1%7D%7B125%7D+)
![\displaystyle 5^{4-3x+8x-2}=5^{-3}\\ 4-3x+8x-2=-3\\ -3x+8x=2-4-3\\ 5x=-5\\ \boxed{ \bf{x=-1}}](https://tex.z-dn.net/?f=+%5Cdisplaystyle+5%5E%7B4-3x%2B8x-2%7D%3D5%5E%7B-3%7D%5C%5C+4-3x%2B8x-2%3D-3%5C%5C+-3x%2B8x%3D2-4-3%5C%5C+5x%3D-5%5C%5C+%5Cboxed%7B+%5Cbf%7Bx%3D-1%7D%7D++)
<u>Проверка</u>:
![\displaystyle 5^{4+3} \cdot 5^{-8-2}=\frac{1}{125}\\ 5^{4+3-8-2}= \frac{1}{125}\\ 5^{-3}= \frac{1}{125}\\ \frac{1}{5^3}= \frac{1}{125}\\ \frac{1}{125}= \frac{1}{125}](https://tex.z-dn.net/?f=+%5Cdisplaystyle+5%5E%7B4%2B3%7D+%5Ccdot+5%5E%7B-8-2%7D%3D%5Cfrac%7B1%7D%7B125%7D%5C%5C+5%5E%7B4%2B3-8-2%7D%3D+%5Cfrac%7B1%7D%7B125%7D%5C%5C+5%5E%7B-3%7D%3D+%5Cfrac%7B1%7D%7B125%7D%5C%5C+%5Cfrac%7B1%7D%7B5%5E3%7D%3D+%5Cfrac%7B1%7D%7B125%7D%5C%5C+%5Cfrac%7B1%7D%7B125%7D%3D+%5Cfrac%7B1%7D%7B125%7D+)
Ответ: x=-1
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
1) 3*4=12 (было слив первоначально). Ответ: 12. 2) Лишнее условие "Маша съела треть слив". 3) Новый вопрос: сколько слив съела Маша? Решение: 3*4=12; 12:3=4. Ответ: :.