Запишем функцию в виде f(x)=1+x^(1/2)/(1-x^(1/2)), тогда f'(x)=(-1/2*x^(-1/2)*(1-x^(1/2))-x^(1/2)*(-1/2*x^(-1/2))/(1-x^(1/2))²=(1/(2*√x))/(1-√x)², тогда при х=4 f'(4)=(1/(2*2))/(1-2)²=1/4
{x-2y-5=0
{x-3z+8=0
N1{1;-2;0},N2{1;0;-3}
a{l;m;n}
l=|-2 0|
|0 -3|=2*3-0*0=6
m=|0 1|
|-3 1|=0*1-1*(-3)=3
n=|1 -2|
|1 0|=1*0-)-2)*1=2
cosα=l/√(l²+m²+n²)=6/√(36+9+4)=6/7⇒α=arccos6/7
cosβ=m/√(l²+m²+n²)=3/7⇒β=arccos3/7
cosγ=n/√(l²+m²+n²)==2/7⇒γ=arccos2/7