У-х=-5
<span>х2-2ху-у2=17 </span>
у=х-5
<span>х^2-2х(х-5)-(х-5)^2=17 </span>
у=х-5
<span>х^2-2х^2+10x-х^2+10x-25-17=0 </span>
у=х-5
<span>-2х^2+20x-42=0 </span>
D=16
x1=7
y1=2
x2=3
<span>y2=-2</span>
1 1/12 * 3 2/3 - (2 5/6 + 3 5/6 * 7/23)*3/5 = 13/12 x 11/3 - (17/6 + 23/6 x 7/23)*3/5 =
(13 * 11)/(12 * 3) - [17/6 + (23 * 7)/(6 * 23)]*3/5 = 143/36 - (17/6 + 161/138)*3/5 =
143/36 - (17*23 + 161)/138*3/5 = 143/36 - (391+161)/238 * 3/5 = 143/36 - 552/238 * 3/5 =
143/36 - 552*3/238*5 = 143/36 -1656/1190 =143/36 - 828/595 = (143*595 - 828*36)/21420 =
(85085 -29808)/21420 = 55277/21420 = 2 12437/21420 = 2,58
2 4/7(4 2/9 + 1 2/9) = 2 4/7 * 5 4/9 = 18/7 * 49/9 = 14
--------------------------------------------
6 2/3(2 5/6 + 3/10 - 1 1/3) = 6 2/3 (( 5 - 1)+ (5/6+3/10-1/3)) =
= 20/3 * (4 +25/30+9/30-10/30) = 20/3 * 4 24/30 = 20/3 * 4 4/5 =
= 20/3 * 24/5 = 32
Координаты точки С, делящей отрезок АВ в отношении |АС|:|СВ|=k
находим по формулам:
![x_C= \frac{x_A+k\cdot x_B}{1+k}, \\ y_C= \frac{y_A+k\cdot y_B}{1+k}, \\ z_C= \frac{z_A+k\cdot z_B}{1+k}](https://tex.z-dn.net/?f=x_C%3D+%5Cfrac%7Bx_A%2Bk%5Ccdot+x_B%7D%7B1%2Bk%7D%2C++%5C%5C+y_C%3D+%5Cfrac%7By_A%2Bk%5Ccdot+y_B%7D%7B1%2Bk%7D%2C+%5C%5C+z_C%3D+%5Cfrac%7Bz_A%2Bk%5Ccdot+z_B%7D%7B1%2Bk%7D+)
Так как по условию задачи k=2, то
![2= \frac{x_A+2\cdot 4}{1+2}\Rightarrow 6=x_A+8\Rightarrow x_A=-2,\\ 3= \frac{y_A+2\cdot 2}{1+2}\Rightarrow 9=y_A+4\Rightarrow y_A=5, \\ -6= \frac{z_A+2\cdot (-3)}{1+2}\Rightarrow -18=z_A-6\Rightarrow z_A=-12,](https://tex.z-dn.net/?f=2%3D+%5Cfrac%7Bx_A%2B2%5Ccdot+4%7D%7B1%2B2%7D%5CRightarrow+6%3Dx_A%2B8%5CRightarrow+x_A%3D-2%2C%5C%5C+3%3D+%5Cfrac%7By_A%2B2%5Ccdot+2%7D%7B1%2B2%7D%5CRightarrow+9%3Dy_A%2B4%5CRightarrow+y_A%3D5%2C+%5C%5C+-6%3D+%5Cfrac%7Bz_A%2B2%5Ccdot+%28-3%29%7D%7B1%2B2%7D%5CRightarrow+-18%3Dz_A-6%5CRightarrow+z_A%3D-12%2C+)
Ответ. А(-2; 5; -12)