Они пересекутся в точке (2;1)
(2y/(y+b)+(b-y)/y)/((b²+y²)/(b+y))=((2y²+b²-y²)/(y*(y+b))/((b²+y²)/(b+y)=
=((b²+y²)/(y*(b+y))/((b²+y²)/(b+y))=1/y.
Y=(x+1)/(x²+3)=0
y`=((x+1)`*(x²+3)-(x+1)*(x²+3)`)/(x²+3)²=0
(1*(x²+3)-(x+1)*2x)/(x²+3)²=0
(x²+3-2x²-2x)/(x²+3)²=0
-(x²+2x-3)/(x²+3)=0 |×(-1)
(x²+2x-3)/(x²+3)=0
x²+3>0 ⇒
x²+2x-3=0 D=16
Ответ: x₁=1 x₂=-3.
√3*sin(2t)+sin(17π/2-t)= t=π/3
=√3*sin(2π/3)+sin(16π/2+π/2-π/3))=√3*sin120°+sin(8π+π/6)=
=√3*√3/2+sin(4*2πn+π/6)=3/2+sin(π/6)=3/2+1/2=4/2=2.
<span>-(3y-5c)+2(y+c)=-3y+5c+2y+2c=7c-y
<span>4a+3c-3a-4c=a-c
</span><span>3(3k-c)-(3c-k)=9k-3c-3c+k=10k
</span><span>4(y-a)-5(2a-y)=4y-4a-10a+5y=9y-14a
</span><span>2a-y-3a+2y=y-a
</span><span>4(a-x)-2(3a-x)=4a-4x-6a+2x=-2a-2x
</span>-(3a-c)+3(3c-a)=-3a+c+9c-3a=10c-6a
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