Comments by "Fhf Fhf" (@fhffhff) on "Подкаст Глеба Соломина"
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y^(1/(y+2))=x^(1/x²) e^(lny/(y+2))(1/y/(y+ 2)-lny(y+2)-²)=0 1+2/y-lny=0 y=4,3+*->y max=4,3^(1/6,3) e^(lnx/x²)(1-2lnx)x-³=0 x=e^0,5(•->x max=e^(0,5/e) 4,3^(1/6,3) >e^(0,5/e) y=1,x=1; y=2,x=2;(-2;-2){(1;1)(2;2)(&2;-2)}
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$√2/2(-cos(π/4-x)+1,5/cos(π/4-x))/cos³x dx=-0,5tgx-3$1/(cos2x+1)/cos2xdx+1,5 ln|u-√0,5|+1,5ln|u+√0,5|-3ln|u|+0,75u-²+u ²/2=-0,5tgx-3ctg(x-π/4)/2/cos2x-3/8sin ²(x-π/4)+3/4ln|cos(x-π/4)|-3/4ln|sin(x-π/4)|+1,5ln|cosx-√0,5|+1,5ln|cosx+√0,5|-3ln|cosx|+1,25/cos²x+c
cosx=u x=±arccosu dx=-+1/√(1-u²)du
0,75=A(u⁴+√0,5u³)+B(u⁴-√0,5u³)+C(u⁴-0,5 u²)+D(u³-0,5u)+E(u²-0,5){B=1,5,A=1,5, C=- 3,D=0,E=-1,5;
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