Comments by "Fhf Fhf" (@fhffhff) on "MyGap"
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(24-((17,5^2-R^2)(-2,5^2+R^2))^0,5/R)/R x=2(-432+R^2+24((17,5^2-R^2)(-2,5^2+R^2))^0,5/R)^0,5,R[2,5;17,5]
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log²2|2x|-5log2|2x|+2|x|log2|2x|-4|x|+6≥0,{|2x|>0,x≠0.(log2|2x|-2,5+|x|)²-(|x|-0,5)²≤0,(log2|2x|-2)(log2|2x|-3+2|x|)≤0,+*-2-*-1+ *1 -2*+>x
log2|2x|-3+2|x|=0,-∞/> 1-3+2=0,log2(-2x) -3-2x=0,\>1-3+2=0,хє[-2;-1]U[1;2],x≠0,хє[-2;-1]U[1;2]
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0,01,0,99*0,02,..,0,99*0,98*..*(1-(n-1)/100)*n/100,.. 99!/(100-n)!*n*0,01^n
99!(100-n)!-¹(50(n-100)-¹-nln(100-n)+ 1,5+n*ln0,01)0,01^n=0->n max=99!/90!*10-¹⁹ Ответ:10-й.
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{(х+у)(х²+у²)=65,(х-у)(х²-у²)=5;{х³+ху²+х² у+у³=65,х³-ху²-х²у+у³=5;{х³+у³=35,ху²+х²у =30;{(х+у)(х²-ху+у²)=35,Ху(х+у)=30;{(х+у) ³=125,Ху(х+у)=30;{х+у=5,Ху=6;{у=5-х,5х -х²=6;-х²+5х-6=0,х=(-6±√(25-4*-1*-6))/-2,[х=2,х=3;[у=3,у=2;
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x=(1±√(4y²-3))y/2/(y²-1)≤(y/2+y²)/(y²-1)= 1+(0,5y+1)/(y²-1)(-0,5y²-0,5-2y)/(y²-1)²=0, y=-2±√3-+>y max=1-0,25(√3+2)=-0,2 5√3+0,5=0,..1+√3/4-0,5=√3/4+0,5=0,8..\> (y/2-√(4y²-3)y/2)/(y²-1)'=-0,5((y²+1)√(4y ²-3)-5y²+3)/(y²-1)²=0,(y²+1)√(4y²-3)=5y²-3,(y²+1)²(4y²-3)=(5y²-3)²,y≥√(3/5),y≤-√0,6; 4y⁶-3y⁴+8y⁴-6y²+4y²-3-25y⁴+30y²-9=0,4y⁶-20y⁴+28y²-9=0,(y²-5/3)³-4/3(y²-5/3)+50 3/108=0,y²-5/3=(-503/216+√(503²/216²+(-4/3)³/27))^(1/3)+(-503/216-√(439*7)/24)^(1/3),y=±√(5/3+(-503/216+√(503²/216²+(-4/3)³/27))^(1/3)+(-503-9√(439* 7))^(1/3)/6)*-1+*1->y max=0/0=(0,5-(4y ²-3)^-0,5*2y²-(4y²-3)⁰,⁵*0,5)/2/y=(0,5-2*1 -0,5)/2=-2/2=-1,min=0/0=(0,5-2-0,5)/2/-1 =-2/-2=1 учитывая значения экстремумов функции,получаем, что других корней нет.
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ab(a+b+2)/(ab-1) b(ba²-2a-b-2)/(ab-1)²= 0,a=(2±√(4-4b(-b-2)))/2/b=(1±√(b²+2b+ 1))/b[a=(b+2)/b,a=-1;+*+>a min=(b+2)/b*b(1+2/b+b+2)/((b+2)/b*b-1)=(b+2)(b²+3b+2)/b*(b+2-1)=(b+2)(b+1)(b+2)/b*(b+1)=b³+13b+4/b+6b²+12,3b²+13+4 b^-2+12b=0,3b⁴+12b³+13b²-4=0,b=-1,3b³+9b²+4b-4=0,(b+1)³-5/3(b+1)-2/3=0,b+ 1=(1/3+√(1/9+(-5/3)³/27))^(1/3)+(1/3- √(-44/3⁶))^(1/3),b=-1+2√(5/9)cos(arccos(9/√(125))/3+2πn/3),+*-1-*-√3/2;-0,5+ *0;-0,5-*√3/2;1+>b min=(-1+2√(5/9)co s(arccos(9/√(125))/3)³+13(-1+2√(5/ 9)cos(arccos(9/√(125))/3)+4/(-1+2√ (5/9)cos(arccos(9/√(125))/3)+6(-1 +2√(5/9)cos(arccos(9/√(125))/3)²+12
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log(-x²+4x-3)(√3sin(π/x)+cos(π/x))=0,{-x²+4x-3>0,-x²+4x-3≠1,√3sin(π/x)+cos(π/x)>0;{(x-(-4+√(16-4*-1*-3))/-2)(x-(-4-√4)/ -2)<0+°-1°3+>x,-x²+4x-4≠0,sin(π/x+π/6)>0;{xє(1;3),х≠(-4±√(16-4*-1*-4))/-2,π/х+π/6є(2πn;π+2πn);{хє(1;3),х≠2,хє(1/(5 /6+2n);1/(-1/6+2n));хє(1;2)U(2;3).2sin(π/x+π/6)=1,sin(π/x+π/6)=1/2,[π/x+π/6=π/6+2πn,π/x+π/6=5/6π+2πn;[x=1/2/n,x=1/(2/3+2n);[0/,0/;0/
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@Vaska.Uticha -0,5x²+0,5ax+9>0 (x-(0,5a-√(0,25a²+18))(x-(0,5a+√(0,25a ²+18)))<0,x(0,5a-√(0,25a²+18);0,5a+√(0, 25a²+18)),x[0;a]{0,5a<√(0,25a²+18), √(0,25a²+18)<0,5a;0/.
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