Comments by "Fhf Fhf" (@fhffhff) on "Продолжение следует"
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x¹/³=u x=u³ dz=3u²du $(0;∞)3u³/(1+u²)/(u²-√3u+1)/(u²+√3u+1)du=-0,5ln|1+u²|+0,25ln|u²-√3u+1|+0,25√3arctg((u-√3/2)/0,5)/0,5+0,25ln|u²+√3u+1|-0,25√3arctg ((u+√3/2)/0,5)/0,5(0;∞)=-∞+∞+∞+√3* π/3=-∞
3u³=A(2u⁵-2u³+2u)+B(u⁴-u²+1)+C(2u⁵+√3 u⁴+u³-u-√3)+D(u⁴+√3u³+2u²+√3u+1)+E (2u⁵-√3u⁴+u³-u+√3)+F(u⁴-√3u³+2u²-√3u+1) {A=-0,5, B=0, C=0,25,D=0,25√3, E=0,25, F=-0,25√3;
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2:3*b√(0,5sin²_b:sin²(_a+_b)+0,5-0,2 5sin²_a:sin²(_a+_b)),2:3b√(0,5+0,5sin²_a:sin²(_a+_b)-0,25sin²_b:sin²(_a+_b)) h=S∆*2/b=b√((1/3√(0,5sin²_b:sin²(_a+_b)+0,5-0,25sin²_a:sin²(_a+_b))+1/3√(0,5+0,5sin²_a:sin²(_a+_ b)-0,25sin²_b:sin²(_a+_b))+1/2)(-1/3√(0,5sin²_b: sin²(_a+_b)+0,5-0,25sin²_a:sin²(_a+ b))+1/3√(0,5+0,5sin²_a:sin²(_a+ b)-0,25sin²_b:sin²(_a+_b))+1/2)(1/3√(0,5sin²_b: sin²(_a+_b)+0,5-0,25sin²_a:sin²(_a+ b))-1/3√(0,5+0,5sin²_a:sin²(_a+ b)-0,25sin²_b:sin²(_a+_b))+1/2)(1/3√(0,5sin²_b: sin²(_a+_b)+0,5-0,25sin²_a:sin²(_a+ b))+1/3√(0,5+0,5sin²_a:sin²(_a+ b)-0,25sin²_b:sin²(_a+_b))-1/2))*2
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x²-|3x-2|≥0 [x≥2/3,(x-1)(x-2)≥0+*+>x;{x<2/3,(x+1,5-√4,25)(x+1,5+√4,25)≥0+* +>x;[{x≥2/3,x(-∞;1]U[2;∞);{x<2/3,x(-∞;-1, 5-√4,25]U[-1,5+√4,25;∞);[x[2/3;1]U[2;∞),x (-∞;-1,5-√4,25]U[-1,5+√4,25;2/3);x(-∞;-1,5-√4,25]U[-1,5+√4,25;1]U[2;∞)
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A(1;4),B(xB;5-xB),C(xC;7xC-3)(xB-1)²+(1-xB)2=(xC-1)²+(7xC-7)² xB=±5(xC-1)+1 10=1,6(xB-1)² xB=1±2,5=3,5;-1,5 xC=1,5;0,5;0,5;1,5 y=(7xC-8+xB)/(xC-xB)(x-xB)+yB=-3x +12;1/3x+2 5/6;-3x+2;1/3x+7
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√(a²-9)+a=6 a[3;6] a²-9=a²-12a+36 a=15/4 {a1²+b²+14/25a1b=36,c1=a1*2/√5 ±√(225/16-a1²/5),c2=-a1+c1*4/√5, c3=1,2a1/√5±√(225/16-a1²/5)*8/5 ±(-4/5/√5a1±3/5√(225/16-a1²/5)), b=0,48a1±√(225/16-a1²/5)*16/5/√ 5±(-8/25a1±6/5/√5√(225/16-a1²/5))±√(225/16-(1,2a1/√5±√(225/16- a1²/5)*8/5±(-4/5/√5a1±3/5√(225/ 16-a1²/5)))²/5),a=0,5arccos(3/5); a1=15/4,b=15/4 c3 c -,c1 c + (15/4=6,15×;3,75✓) c1=3√5(+) R=15/4/2/√(1/5)=15/8√5
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{b=(5-ax)/y,y=(10-ax²)/(5-ax), x=(6a²+6 15a+300)(7a-15)/a/(101a³-564a²+595a+ 75),[a=0,a=-205,12,a=-50,6,a=-1,76,a=-0,49, a=4,63*10-⁵,a=2,14; {(-205,12;2,29;-0,001; 2,09)(-50,6;2,90;0,0077;1,86)(-1,76;3,91; 3,63;2,91)(-0,49;2,50;-0,0094;2)(4,63*10-⁵; -5,43*10-⁵;-1295515,49;-1195'833,26)(2,14;2,37;0,063;2,05)} ax⁵+by⁵=91,32;64, 56;-293,38;80;3,02*10²⁶;85,81
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$√(x²+x+1)/(x+1)dx-х+ln|x+1|+c
√(x²+x+1)=u,x=-0,5±√(u²-0,75) dx=±(u²-0,75)-⁰,⁵udu
$(±0,5(u²-0,75)⁰,⁵±0,375(u²-0,75)-⁰, ⁵)/(1-u²)du+u+0,5ln|u-1|-0,5ln|u+1|
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