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
Х(х+у+1) = 10
у(у+х+1) = 20
и заметить, что выражения в скобках получились одинаковые...
х+у+1 = 10/х ---подставим во второе уравнение...
у*10/х = 20
у/х = 2
у = 2х ---а теперь это подставим в первое уравнение...
х(х+2х+1) = 10
х(3х+1) = 10
3x^2 + x - 10 = 0
D = 1+4*3*10 = 11^2
x1 = (-1-11)/6 = -2
x2 = (-1+11)/6 = 5/3
y1 = -4
y2 = 10/3
Ответ: (-2; -4), (1_2/3; 3_1/3)
Task/26898605
-------------------
а)
x⁴<span>- 5x</span>² +4 ≤ 0 ⇔(x²-4)(x²-1) ≤ 0 ⇔(x+2)(x+1)(x-1)(x-2) ≤ 0
+ - + - +
---------- [ -2] /////////////[-1] ------------[1] //////////////[2] -----------
x∈ [ -2; -1 ] ∪ [ 1 ; 2] .
-------
б)
2x⁴ +x² - 3 >0 ⇔2(x² +3/2)(x² -1) >0 ⇔x² -1 >0⇔(x+1)(x-1) >0
+ - +
///////////////////////////(-1) ------------(1) ///////////////////////////
x∈ (-∞; -1 ) ∪ (1 ; + ∞) .
-------
в)
5x⁴ - 9x² +8 ≥ 0⇔5( x² -9/10)² + 79 /20 ≥ 0 * * * или D=9² -160 < 0 * * *
x (-∞ ; + ∞) .
-------
г)
-6x⁴ - 7x² +10 < 0⇔6x⁴ + 7x² -10 > 0⇔6(x²+2)(x² -5/6 )> 0⇔x² -5/6> 0⇔
(x+ √30 /6) (x -√30 /6) >0
+ - +
/////////////////////////// - (√30) /6 ------------(√30) /6) ///////////////////////////
x∈ (-∞; - (√30) /6 ) ∪ ( (√30) / 6 ; + ∞) .
10х-7-32=-16
10х=-16+7+32
10х=23
х=23:10
х=2,3
(17*2-1/32)*24/11=33/32*24/11=9/4=2,25