Пусть x - скорость нагнетания воды первым насосом, пусть y - скорость нагнетания воды вторым насосом, пусть t - время. V - объём бассейна.
(x+y)*t=V; t=48/60=0,8.
xt1=y(t1+1/3)=(x+y)4/5;=> xt1=yt1+y/3=4x/5+4y/5;
![\left \{ {{xt1=4x/5+4y/5} \atop {yt1+y/3=4x/5+4y/5;}} \right. => \left \{ {{5xt1=4x+4y} \atop {5yt1+5y/3=4x+4y;}} \right. => \left \{ {{t1=\frac{4x+4y}{5x}} \atop {5yt1=4x+7y/3;}} \right.](https://tex.z-dn.net/?f=+%5Cleft+%5C%7B+%7B%7Bxt1%3D4x%2F5%2B4y%2F5%7D+%5Catop+%7Byt1%2By%2F3%3D4x%2F5%2B4y%2F5%3B%7D%7D+%5Cright.+%3D%3E+%5Cleft+%5C%7B+%7B%7B5xt1%3D4x%2B4y%7D+%5Catop+%7B5yt1%2B5y%2F3%3D4x%2B4y%3B%7D%7D+%5Cright.+%3D%3E+%5Cleft+%5C%7B+%7B%7Bt1%3D%5Cfrac%7B4x%2B4y%7D%7B5x%7D%7D+%5Catop+%7B5yt1%3D4x%2B7y%2F3%3B%7D%7D+%5Cright.+)
![\left \{ {{t1=\frac{4x+4y}{5x}} \atop {t1=\frac{12x+7y;}{15y}}} \right. =>\frac{4x+4y}{5x}=\frac{12x+7y;}{15y} => (4x+4y)15y=(12x+7y)5x](https://tex.z-dn.net/?f=%5Cleft+%5C%7B+%7B%7Bt1%3D%5Cfrac%7B4x%2B4y%7D%7B5x%7D%7D+%5Catop+%7Bt1%3D%5Cfrac%7B12x%2B7y%3B%7D%7B15y%7D%7D%7D+%5Cright.+%3D%3E%5Cfrac%7B4x%2B4y%7D%7B5x%7D%3D%5Cfrac%7B12x%2B7y%3B%7D%7B15y%7D+%3D%3E+%284x%2B4y%2915y%3D%2812x%2B7y%295x+)
(1)
t1(x-y)=y/3; => ![t1=\frac{y}{3(x-y)}](https://tex.z-dn.net/?f=t1%3D%5Cfrac%7By%7D%7B3%28x-y%29%7D)
![\left \{ {{t1=\frac{4x+4y}{5x}} \atop {t1=\frac{y}{3(x-y)}} \right. => \frac{4x+4y}{5x}=\frac{y}{3(x-y)}](https://tex.z-dn.net/?f=%5Cleft+%5C%7B+%7B%7Bt1%3D%5Cfrac%7B4x%2B4y%7D%7B5x%7D%7D+%5Catop+%7Bt1%3D%5Cfrac%7By%7D%7B3%28x-y%29%7D%7D+%5Cright.+%3D%3E+%5Cfrac%7B4x%2B4y%7D%7B5x%7D%3D%5Cfrac%7By%7D%7B3%28x-y%29%7D%C2%A0)
5xy=3(4x+4y)(x-y)=> 5xy=(4x+4y)(3x-3y)=>5xy=12(x*x-y*y)=> 12x*x-5xy-12y*y=0; (2)
(1)+(2):
24x*x-30xy=0 => 24x=30y=> 4xx=5y; => x=5y/4;
t1*5y/4=y(t1+1/3) => 5t1/4=t1+1/3 => 15t1=12t1+4 => 3t1=4 => t1=4/3
Ответ: 4/3 часа
Х*(х-у)=3
у*(х-у)=2
х-у=3/х
х-у=2/у
3/х=2/у
3у=2х
х=3/2у=1.5у
у*(1.5у-у)=2
0.5у^2=2
у^2=2/0.5=4
у1=-1 => х1=1.5*1=1.5
у2=1 => х2=1.5*(-1)=-1.5
<span>б)36+95-205*48/164=36+95-60=71</span>