Решите пример
[tex] \frac{y + 3}{2y + 2} - \frac{y + 1}{2y - 2} + \frac{3}{ {y}^{2} - 1} [/tex]
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[tex]\frac{y+3}{2y+2}-\frac{y+1}{2y-2}+\frac{3}{y^2-1}=\frac{y+3}{2(y+1)}-\frac{y+1}{2(y-1)}+\frac{3}{(y-1)(y+1)}==\frac{(y+3)(y-1)-(y+1)^2+3\cdot 2}{2(y-1)(y+1)}=\frac{y^2+2y-3-y^2-2y-1+6}{2(y-1)(y+1)}=\frac{2}{2(y-1)(y+1)}=\frac{1}{y^2-1}[/tex]

[tex]\frac{y+3}{2y+2}-\frac{y+1}{2y-2}+\frac{3}{y^{2}-1}=\frac{y+3}{2(y+1)}-\frac{y+1}{2(y-1)}+\frac{3}{(y-1)(y+1)}=\frac{y^{2} -y+3y-3-y^{2}-2y-1+6 }{2(y-1)(y+1)}=\frac{2}{2(y^{2}-1)}=\frac{1}{y^{2} -1}[/tex]

первое число - x, второе - y

x-0.1y=0.5(y+0.5x)

x-0.1y=0.5y+0.25x

x-0.25x=0.5y+0.1y

0.75x=0.6y

75x=60y

5x=4y

x/y=4/5

ответ: 4/5

0,8x^2-19.8x-5=?

d=(-19,8)^2-4*0,8*(-5)=392? 04+16=408,04

sqr(d)=20,2

х1= (19,8+20,2)/(2*0,8)=40/1.6=25

х2=(19,8-20,2)/(2*0,8)=-0,4/1,6=-0,25

0,8x^2-19.8x-5=0,8(х-25)(х+0,25)