Ищем точку пересечения с осью Ох (y=0)
y=0
y=7x-35
7x-35=0;
7x=35;
x=35:7;
x=5
(5;0)
Ищем точку пересечения с осью Оy (x=0)
x=0
y=7x-35=7*0-35=-35
(-35;0)
5х-3у=11, 2х-4у=3
10х-6у=22, 10х-20у=15
10х-6у-10х+20у=22-15
14у=7
у=1/2
5х-3×1/2=11, 2х-4×1/2=3
5х-1,5=11, 2х-2=3
5х=12,5, 2х=5
х=2,5, х=2,5.
Ответ: (2,5;0,5).
![\frac{x-5}{x+6}- \frac{x+6}{x-5}=0](https://tex.z-dn.net/?f=+%5Cfrac%7Bx-5%7D%7Bx%2B6%7D-++%5Cfrac%7Bx%2B6%7D%7Bx-5%7D%3D0)
![\frac{(x-5)(x-5)-(x+6)(x+6)}{(x+6)(x-5)} = \frac{(x-5)^{2}-(x+6)^2}{(x+6)(x-5)}= \frac{x^2-10x+25-x^2-12x-36}{(x+6)(x-5)}=](https://tex.z-dn.net/?f=++%5Cfrac%7B%28x-5%29%28x-5%29-%28x%2B6%29%28x%2B6%29%7D%7B%28x%2B6%29%28x-5%29%7D+%3D++%5Cfrac%7B%28x-5%29%5E%7B2%7D-%28x%2B6%29%5E2%7D%7B%28x%2B6%29%28x-5%29%7D%3D+%5Cfrac%7Bx%5E2-10x%2B25-x%5E2-12x-36%7D%7B%28x%2B6%29%28x-5%29%7D%3D)
![= \frac{-22x-11}{x^2-5x+6x-30}= \frac{-22x-11}{x^2+x-30}=0](https://tex.z-dn.net/?f=%3D+%5Cfrac%7B-22x-11%7D%7Bx%5E2-5x%2B6x-30%7D%3D+%5Cfrac%7B-22x-11%7D%7Bx%5E2%2Bx-30%7D%3D0)
Дробь будет равна нулю если числитель будет равен нулю (знаменатель не может быть равен нулю, на ноль делить нельзя)
-22x-11=0
-22x=11
x=-11/22=-1/2
![\frac{2}{x-1}+3= \frac{1}{x+1}](https://tex.z-dn.net/?f=+%5Cfrac%7B2%7D%7Bx-1%7D%2B3%3D+%5Cfrac%7B1%7D%7Bx%2B1%7D)
![\frac{2}{x-1}+3- \frac{1}{x+1}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B2%7D%7Bx-1%7D%2B3-+%5Cfrac%7B1%7D%7Bx%2B1%7D%3D0)
Приводим к общему знаменателю, который представляет из себя разность квадратов (x²-1)
![\frac{2(x+1)+3(x^2-1)-(x-1)}{x^2-1}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B2%28x%2B1%29%2B3%28x%5E2-1%29-%28x-1%29%7D%7Bx%5E2-1%7D%3D0)
![\frac{2x+2+3x^2-3-x+1}{x^2-1}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B2x%2B2%2B3x%5E2-3-x%2B1%7D%7Bx%5E2-1%7D%3D0)
![\frac{3x^2+x}{x^2-1}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B3x%5E2%2Bx%7D%7Bx%5E2-1%7D%3D0)
3x²+x=0
x(3x+1)=0
x=0 3x+1=0
3x=-1
x=-1/3
Получилось два корня 0 и =-1/3
![\frac{1-2x}{x^2-1}- \frac{1}{x+1} = \frac{2}{x-1}](https://tex.z-dn.net/?f=+%5Cfrac%7B1-2x%7D%7Bx%5E2-1%7D-+%5Cfrac%7B1%7D%7Bx%2B1%7D+%3D+%5Cfrac%7B2%7D%7Bx-1%7D+)
![\frac{1-2x}{x^2-1}- \frac{1}{x+1}- \frac{2}{x-1}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B1-2x%7D%7Bx%5E2-1%7D-+%5Cfrac%7B1%7D%7Bx%2B1%7D-+%5Cfrac%7B2%7D%7Bx-1%7D%3D0+)
Приводим к общему знаменателю, им будет (x²-1) - разность квадратов
![\frac{1-2x-(x-1)-2(x+1)}{x^2-1}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B1-2x-%28x-1%29-2%28x%2B1%29%7D%7Bx%5E2-1%7D%3D0)
![\frac{1-2x-x+1-2x-1}{x^2-1}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B1-2x-x%2B1-2x-1%7D%7Bx%5E2-1%7D%3D0)
![\frac{1-5x}{x^2-1}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B1-5x%7D%7Bx%5E2-1%7D%3D0)
1-5x=0
-5x=-1
x=1/5
![4 \frac{1}{4}- \frac{x+1}{x-2}= \frac{x-2}{x+1}](https://tex.z-dn.net/?f=4+%5Cfrac%7B1%7D%7B4%7D-+%5Cfrac%7Bx%2B1%7D%7Bx-2%7D%3D+%5Cfrac%7Bx-2%7D%7Bx%2B1%7D+)
<span>
![\frac{17}{4}- \frac{x+1}{x-2}- \frac{x-2}{x+1}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B17%7D%7B4%7D-+%5Cfrac%7Bx%2B1%7D%7Bx-2%7D-+%5Cfrac%7Bx-2%7D%7Bx%2B1%7D%3D0)
Приводим к общему знаменателю, который будет 4(x-2</span>)(x+1)
![\frac{17(x-2)(x+1)-4(x+1)(x+1)-4(x-2)(x-2)}{4(x-2)(x+1)}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B17%28x-2%29%28x%2B1%29-4%28x%2B1%29%28x%2B1%29-4%28x-2%29%28x-2%29%7D%7B4%28x-2%29%28x%2B1%29%7D%3D0)
![\frac{17x^2+17x-34x-34-4x^2-4x-4x-4-4x^2+8x+8x-16}{4(x-2)(x+1)}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B17x%5E2%2B17x-34x-34-4x%5E2-4x-4x-4-4x%5E2%2B8x%2B8x-16%7D%7B4%28x-2%29%28x%2B1%29%7D%3D0)
![\frac{9x^2-9x-54}{4(x-2)(x+1)}=0](https://tex.z-dn.net/?f=+%5Cfrac%7B9x%5E2-9x-54%7D%7B4%28x-2%29%28x%2B1%29%7D%3D0)
9x²-9x-54=0
D=(-9)²-4*9*(-54)=81+1944=2025
![x_{1}= \frac{9- \sqrt{2025}}{2*9}= \frac{9-45}{18}=-2](https://tex.z-dn.net/?f=+x_%7B1%7D%3D+%5Cfrac%7B9-+%5Csqrt%7B2025%7D%7D%7B2%2A9%7D%3D+%5Cfrac%7B9-45%7D%7B18%7D%3D-2)
![x_{2}= \frac{9+45}{18}=3](https://tex.z-dn.net/?f=+x_%7B2%7D%3D+%5Cfrac%7B9%2B45%7D%7B18%7D%3D3)
<u>Способ 1</u>. ctg²α = 1 / tg²α.
![\displaystyle \frac{1}{1+ ctg^2 \alpha}+ \frac{1}{1+tg^2 \alpha}= \frac{1}{1+ \frac{1}{tg^2 \alpha}}+\frac{1}{1+ tg^2 \alpha}= \frac{1}{\frac{tg^2 \alpha +1}{tg^2 \alpha}}+ \frac{1}{1+ tg^2 \alpha}=](https://tex.z-dn.net/?f=+%5Cdisplaystyle+%5Cfrac%7B1%7D%7B1%2B+ctg%5E2+%5Calpha%7D%2B+%5Cfrac%7B1%7D%7B1%2Btg%5E2+%5Calpha%7D%3D+%5Cfrac%7B1%7D%7B1%2B+%5Cfrac%7B1%7D%7Btg%5E2+%5Calpha%7D%7D%2B%5Cfrac%7B1%7D%7B1%2B+tg%5E2+%5Calpha%7D%3D+%5Cfrac%7B1%7D%7B%5Cfrac%7Btg%5E2+%5Calpha+%2B1%7D%7Btg%5E2+%5Calpha%7D%7D%2B+%5Cfrac%7B1%7D%7B1%2B+tg%5E2+%5Calpha%7D%3D+)
![\displaystyle = \frac{tg^2 \alpha}{tg^2 \alpha +1}+ \frac{1}{1+ tg^2 \alpha} = \frac{tg^2 \alpha +1}{tg^2 \alpha +1}=1](https://tex.z-dn.net/?f=+%5Cdisplaystyle+%3D+%5Cfrac%7Btg%5E2+%5Calpha%7D%7Btg%5E2+%5Calpha+%2B1%7D%2B+%5Cfrac%7B1%7D%7B1%2B+tg%5E2+%5Calpha%7D+%3D+%5Cfrac%7Btg%5E2+%5Calpha+%2B1%7D%7Btg%5E2+%5Calpha+%2B1%7D%3D1+)
Ответ: 1.
<u>Способ 2</u>. sin²α+cos²α=1; tg²α=sin²α/cos²α; ctg²α=cos²α/sin²α.
![\displaystyle \frac{1}{1+ ctg^2 \alpha}+ \frac{1}{1+ tg^2 \alpha}=\frac{1}{1+ \frac{cos^2 \alpha}{sin^2 \alpha}}+ \frac{1}{1+ \frac{sin^2 \alpha}{cos^2 \alpha}}=\frac{1}{\frac{sin^2 \alpha+cos^ \alpha}{sin^2 \alpha}}+\\ +\frac{1}{\frac{cos^2 \alpha+sin^2 \alpha}{cos^2 \alpha}}=\frac{sin^2 \alpha}{sin^2 \alpha+cos^2 \alpha}+ \frac{cos^2 \alpha}{cos^2 \alpha+sin^2 \alpha}=sin^2 \alpha+cos^2 \alpha=1](https://tex.z-dn.net/?f=+%5Cdisplaystyle+%5Cfrac%7B1%7D%7B1%2B+ctg%5E2+%5Calpha%7D%2B+%5Cfrac%7B1%7D%7B1%2B+tg%5E2+%5Calpha%7D%3D%5Cfrac%7B1%7D%7B1%2B+%5Cfrac%7Bcos%5E2+%5Calpha%7D%7Bsin%5E2+%5Calpha%7D%7D%2B+%5Cfrac%7B1%7D%7B1%2B+%5Cfrac%7Bsin%5E2+%5Calpha%7D%7Bcos%5E2+%5Calpha%7D%7D%3D%5Cfrac%7B1%7D%7B%5Cfrac%7Bsin%5E2+%5Calpha%2Bcos%5E+%5Calpha%7D%7Bsin%5E2+%5Calpha%7D%7D%2B%5C%5C+%2B%5Cfrac%7B1%7D%7B%5Cfrac%7Bcos%5E2+%5Calpha%2Bsin%5E2+%5Calpha%7D%7Bcos%5E2+%5Calpha%7D%7D%3D%5Cfrac%7Bsin%5E2+%5Calpha%7D%7Bsin%5E2+%5Calpha%2Bcos%5E2+%5Calpha%7D%2B+%5Cfrac%7Bcos%5E2+%5Calpha%7D%7Bcos%5E2+%5Calpha%2Bsin%5E2+%5Calpha%7D%3Dsin%5E2+%5Calpha%2Bcos%5E2+%5Calpha%3D1+)
Ответ: 1.