Comments by "Fhf Fhf" (@fhffhff) on "hrendyabliki"
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2log(3)tgx=log(2)sinx,{tgx>0,sinx>0;{xє(πn;π/2+πn),xє(2πn;π+2πn);xє(2πn;π/2+2πn) log(3)tg²x=log(2)sinx,tg²x=3^log (2)sinx,≤1,xє(2πn;π/4+2πn),3^log(2)(√2/2)=3^-0,5,tgx=3^-0,25,x=arctg3^-0,25+πn, π/4>arctg3^-0,25>33°..x=π/6,3/9=3^log(2)0,5=3^-1=1/3,✓ x=π/6+2πn,2tgx*1/cos²x-3^l0g(2)sinx*ln3*1/sinx/ln2*cosx=0, />,2*√3/3/(3/4)-3^lo g(2)0,5ln3/0,5/ln2*√3/2=8√3/9-1/3*log (2)3*√3/2=√3/3(8/3-log(2)3/2)<0,a значит функция убывает и при хє(2π n;π/6+2πn) решений нет.
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{√x+y=7,x+√y=11;{y=7-√x,x+√(7-√x)=11; √(7-√x)=-x+11,7-√x=x²-22x+121,x≤11,x=(-x²+22x-114)²,x²-22x+114≤0,(x-(22+√(484- 4*114))/2)(x-(11-√(484-456)/2))≤0+*11- √7-11+√7+>x xє[11-√7;11+√7],хє[11-√7; 11]*/> 11-√7>0,11<49,11+√7>0,-(11+√ 7)/((11+√7+38)/√7)+11+√7=-(11√7+7)/(49+√7)+11+√7=-(11√7*49-77+7*49-7√7)/(49²-7)+11+√7=-(532√7+266)/2394+11 +√7=7√7/9+10 8/9,х=12,9..х=7,9..х⁴-44х³ +672х²-5017х+114²|х-12,9..=х³-21,1х²+39 9,81х+140,549,х³-21,1х²+398,81х-140,549|х-7,9=х²-13,2х+294,53 х=(13,2±√(13,2²- 4*294,53))/2,х=6,6±√-1003,88/2 0/х=12,9>11,х=7,9.. у=7-√7,9..=7-(3-0,18)= 4,18..
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{(x+1)²+(y+2)²=25,z²+(w-1)²=144,xw+yz-x +w+2z-61≥0,x≥(-yz-w-2z+61)/(w-1),w-1≥0, w≥1 x=±√(25-(y+2)²)-1,z=±√(144-(w-1)²), ±√(25-(y+2)²)-1≥(-y*±√(144-(w-1)²)-w+61)/(w-1),±√(25-(y+2)²)(w-1)≥-+y√(144-(w- 1)²)-w+61,[{(-y²-4y+21)(w-1)²≥(-y√(-w²+2w +143)-w+61)²,-y√(-w²+2w+143)-w+61≥0, y√(-w²+2w+143)-w+61<0;[{-yw²+2wy-y² 4yw²+8yw-4y+21w²-42w+21≥y²(-w²+2w+143)+w²+3721+2yw√(-w²+2w+143)-122y√(-w²+2w+143)-122w,y√(-w²+2w+143)≤-w+61;[{-y²w²+2y²w+143y²≤w²-122w+37 21,y≥0;{y<0,✓;[{-5yw²-2wy²+w²y²+10wy-144y²+ -4y+20w²+80w-3700≥(2yw-122y)√(-w²+2w+143),y≤-w+6/√(-w²+2w+143);[{-y²w²+2y²w+143y²≤w²-122w+3721,y≥0;{y<0,✓;[{(y-(5w²-10w+4+2(w-61)√(-w²+2w+143)+√((5w²-10w+4+2(w-61)√(-w²+2w+143))²-4+w²-2w-144)(20w²+60w-3700)))/2/(w²-2w-144))(y(-5w²-10w+4+2(w-61)√(-w²+2w+143)-√((5w²-10w+4)²+4(5w²-10w+4)(w-61)√(-w²+2w+143)+4(w-61)²(-w²+2w+143)-80w⁴-80w³+26800w²+5760w-2131200))/2/(w²-2w-144))≤0,
,y≤(-w+61)/√(-w²+2w+143);[{-y²w²+2y²w+143y²≤w²-122w+3721,y≥0;{y<0,✓;
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$(3x³+5x²-25x-1)/(x³-3x+2)dx=$(3+(5x²- 16x-7)/(x-1)/(x²+x-2))dx=$(3+A/(x-1)+B/(x-1)²+C/(x+2))dx=3x-6(x-1)^-1/-1+5ln|x+2|+C
5x²-16x-7=A(x-1)(x+2)+B(x+2)+C(x-1)² =A(x²+x-2)+Bx+2B+Cx²-2Cx+C=Ax²+Ax-2A+Bx+2B+Cx²-2Cx+C,{A+C=5,A+B-2C=- 16,-2A+2B+C=-7|•-0,5;{A+C=5,-B+3C=21,B +1,5C=1,5;{A+C=5,-B+3C=21,4,5C=22,5;{A=0,B=-6,C=5;
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