Решение задания смотри на фотографии
5\8. Пусть х-числитель,тогда х+3 - знаменатель. Составляем уравнение для числителя: 3х-7. Для знаменателя: 2(х+3)-11. Получили дробь. В условии говорится,что это обратная дробь к нашей,поэтому,переворавиваем ее и приравниваем числители. Получаем,что х=5 - это наш числитель,Подставляем 5 в знаменатель,получам 8.Вот и ве)))
Найдем х и у методом сложения, сложив обе части системы:
![\left \{ {{3x-2y=7} \atop {2x+2y=8}} \right. \\ 3x+2x=7+8\\ 5x=15\\ x=3\\ \left \{ {{x=3} \atop {2x+2y=8}} \right. \\ \left \{ {{x=3} \atop {2*3+2y=8}} \right. \\ \left \{ {{x=3} \atop {2y=2}} \right. \\ \left \{ {{x=3} \atop {y=1}} \right. \\ OTBET:(3;1).](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B3x-2y%3D7%7D+%5Catop+%7B2x%2B2y%3D8%7D%7D+%5Cright.+%5C%5C+3x%2B2x%3D7%2B8%5C%5C+5x%3D15%5C%5C+x%3D3%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%3D3%7D+%5Catop+%7B2x%2B2y%3D8%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%3D3%7D+%5Catop+%7B2%2A3%2B2y%3D8%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%3D3%7D+%5Catop+%7B2y%3D2%7D%7D+%5Cright.++%5C%5C++%5Cleft+%5C%7B+%7B%7Bx%3D3%7D+%5Catop+%7By%3D1%7D%7D+%5Cright.++%5C%5C+OTBET%3A%283%3B1%29.)
2/5х + 5/6 = 1/6 - 7/25х
10/25х + 7/25х = 1/6 - 5/6
17/25х = - 2/3
Х = - 2/3 : 17/25
Х = - 50/51
1)
![(\frac{8a}{a^{2}-b^{2}}+ \frac{3}{b-a}-\frac{4}{a+b}): \frac{1}{5a-5b}=(\frac{8a}{(a-b)(a+b)}+ \frac{3}{b-a}-\frac{4}{a+b})](https://tex.z-dn.net/?f=%28%5Cfrac%7B8a%7D%7Ba%5E%7B2%7D-b%5E%7B2%7D%7D%2B+%5Cfrac%7B3%7D%7Bb-a%7D-%5Cfrac%7B4%7D%7Ba%2Bb%7D%29%3A+%5Cfrac%7B1%7D%7B5a-5b%7D%3D%28%5Cfrac%7B8a%7D%7B%28a-b%29%28a%2Bb%29%7D%2B+%5Cfrac%7B3%7D%7Bb-a%7D-%5Cfrac%7B4%7D%7Ba%2Bb%7D%29)
·5(a-b) =
![\frac{8a-3(a+b)-4(a-b)}{(a-b)(a+b)}](https://tex.z-dn.net/?f=+%5Cfrac%7B8a-3%28a%2Bb%29-4%28a-b%29%7D%7B%28a-b%29%28a%2Bb%29%7D+)
·5(a-b) =
![\frac{8a-3a-3b-4a+4b}{a+b}](https://tex.z-dn.net/?f=+%5Cfrac%7B8a-3a-3b-4a%2B4b%7D%7Ba%2Bb%7D+)
·5 =
![\frac{a+b}{a+b}](https://tex.z-dn.net/?f=+%5Cfrac%7Ba%2Bb%7D%7Ba%2Bb%7D+)
·5 = 5
2)
![\frac{2x+3}{x^{2}-2x}- \frac{x-3}{x^{2}+2x}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B2x%2B3%7D%7Bx%5E%7B2%7D-2x%7D-+%5Cfrac%7Bx-3%7D%7Bx%5E%7B2%7D%2B2x%7D%3D0++)
![\frac{2x+3}{x(x-2)}- \frac{x-3}{x(x+2)}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B2x%2B3%7D%7Bx%28x-2%29%7D-+%5Cfrac%7Bx-3%7D%7Bx%28x%2B2%29%7D%3D0+)
![\frac{(2x+3)(x+2)-(x-3)(x-2)}{x(x-2)(x+2)}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B%282x%2B3%29%28x%2B2%29-%28x-3%29%28x-2%29%7D%7Bx%28x-2%29%28x%2B2%29%7D%3D0+)
![\frac{2x^{2}+4x+3x+6-x^{2}+2x+3x-6}{x(x-2)(x+2)}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B2x%5E%7B2%7D%2B4x%2B3x%2B6-x%5E%7B2%7D%2B2x%2B3x-6%7D%7Bx%28x-2%29%28x%2B2%29%7D%3D0+)
![\frac{x^{2}+12x}{x(x-2)(x+2)}=0](https://tex.z-dn.net/?f=+%5Cfrac%7Bx%5E%7B2%7D%2B12x%7D%7Bx%28x-2%29%28x%2B2%29%7D%3D0+)
![\left \{ {{ x^{2} +12x=0} \atop {x(x-2)(x+2) \neq 0}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B+x%5E%7B2%7D+%2B12x%3D0%7D+%5Catop+%7Bx%28x-2%29%28x%2B2%29+%5Cneq+0%7D%7D+%5Cright.+)
![\left \{ {{ x(x+12)=0} \atop {x(x-2)(x+2) \neq 0}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B+x%28x%2B12%29%3D0%7D+%5Catop+%7Bx%28x-2%29%28x%2B2%29+%5Cneq+0%7D%7D+%5Cright.+)
![\left \{ {{ x=-12;0} \atop {x \neq -2;0;2}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7B+x%3D-12%3B0%7D+%5Catop+%7Bx+%5Cneq+-2%3B0%3B2%7D%7D+%5Cright.+)
x = -12
3 a)
![\frac{(x-2)(x+2)}{x-3}<0](https://tex.z-dn.net/?f=+%5Cfrac%7B%28x-2%29%28x%2B2%29%7D%7Bx-3%7D%3C0+)
x є (-oo;-2) U (2;3)
3 б)
![\frac{x^{2}-10x+25}{x^{2}-4x-12}\geq0](https://tex.z-dn.net/?f=+%5Cfrac%7Bx%5E%7B2%7D-10x%2B25%7D%7Bx%5E%7B2%7D-4x-12%7D%5Cgeq0+)
![\frac{(x-5)^{2}}{(x+2)(x-6)}\geq0](https://tex.z-dn.net/?f=+%5Cfrac%7B%28x-5%29%5E%7B2%7D%7D%7B%28x%2B2%29%28x-6%29%7D%5Cgeq0+)
x є (-oo;-2) U (6;+oo)
4 a)
![(\frac{1}{ n^{2}-n }+\frac{1}{ n^{2}+n }):\frac{n+3}{ n^{2}-1}=(\frac{1}{n(n-1)}+\frac{1}{n(n+1)})](https://tex.z-dn.net/?f=%28%5Cfrac%7B1%7D%7B+n%5E%7B2%7D-n+%7D%2B%5Cfrac%7B1%7D%7B+n%5E%7B2%7D%2Bn+%7D%29%3A%5Cfrac%7Bn%2B3%7D%7B+n%5E%7B2%7D-1%7D%3D%28%5Cfrac%7B1%7D%7Bn%28n-1%29%7D%2B%5Cfrac%7B1%7D%7Bn%28n%2B1%29%7D%29)
·
![\frac{(n-1)(n+1)}{n+3}=\frac{n+1+n-1}{n(n-1)(n+1)}](https://tex.z-dn.net/?f=%5Cfrac%7B%28n-1%29%28n%2B1%29%7D%7Bn%2B3%7D%3D%5Cfrac%7Bn%2B1%2Bn-1%7D%7Bn%28n-1%29%28n%2B1%29%7D)
·
![\frac{(n-1)(n+1)}{n+3}=\frac{2}{n+3}](https://tex.z-dn.net/?f=%5Cfrac%7B%28n-1%29%28n%2B1%29%7D%7Bn%2B3%7D%3D%5Cfrac%7B2%7D%7Bn%2B3%7D)
4 б)
![\frac{2}{n+3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7Bn%2B3%7D)
при n = -1
![\frac{2}{-1+3}](https://tex.z-dn.net/?f=%5Cfrac%7B2%7D%7B-1%2B3%7D)
= 1
5 a)
x² + y² - 2x + 4y + 5 ≥ 0
(x² - 2x + 1) + (y² + 4y + 4) ≥ 0
(x - 1)² + (y + 2)² ≥ 0
5 б)
x⁴ - 3x² - 2x + 6 > 0
(x⁴ - 4x² +4) + (x² - 2x + 1) + 1 > 0
(x² - 2)² + (x - 1)² + 1 > 0
5 в)
x² + 2x +
![\frac{1}{ x^{2} +2x+2}](https://tex.z-dn.net/?f=+%5Cfrac%7B1%7D%7B+x%5E%7B2%7D+%2B2x%2B2%7D+)
≥ 0
![\frac{(x^{2} +2x)^{2}+2(x^{2} +2x)+1}{(x^{2} +2x+1)+1}](https://tex.z-dn.net/?f=+%5Cfrac%7B%28x%5E%7B2%7D+%2B2x%29%5E%7B2%7D%2B2%28x%5E%7B2%7D+%2B2x%29%2B1%7D%7B%28x%5E%7B2%7D+%2B2x%2B1%29%2B1%7D+)
≥ 0
![\frac{(x^{2} +2x+1)^{2}}{(x+1)^{2}+1}](https://tex.z-dn.net/?f=+%5Cfrac%7B%28x%5E%7B2%7D+%2B2x%2B1%29%5E%7B2%7D%7D%7B%28x%2B1%29%5E%7B2%7D%2B1%7D+)
≥ 0
6)
x⁴ - x³ - 3x² + 4x - 4 = 0
(x⁴ - x³ + x²) - (4x² - 4x + 4) = 0
x²(x² - x + 1) - 4(x² - x + 1) = 0
(x² - x + 1)(x² - 4) = 0
(x² - x + 1)(x - 2)(x+2) = 0
Трёхчлен x² - x + 1 не расскладывается на множители.
Поэтому х = -2 и 2.
7)
Пусть двузначное число [xy].
Тогда: [xy1] - [1xy] = 234 или:
100x + 10y + 1 - (100 + 10x + y) = 234
90x + 9y = 333
10x + y = 37
Т. е. двузначное число [xy] = 37