Comments by "Fhf Fhf" (@fhffhff) on "ВЫХОД ЕСТЬ!"
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x²+4xy+y²+1/4=0, x=(-4y±√(16y²-4*(y²+ 1/4)))/2=-2y±√(16y²-4y²-1)/2=-2y±√(12y²- 1)/2,12y²-1≥0,(y-1/√12)(y+1/√12)≥0, +**+>y,yє(-∞;-1/√12]U[1/√12;+∞) √(1-4(-2y±√(3y²-1/4))²)-√(1-4y²)-2x -2y=0,√(2-28y²±16y√(3y²-1/4))-√(1-4y²)+2y-+2√(3y²-1/4)=0,2-28y²±16y√(3y²-1/4)-2√(2-28y²±16y√(3y²-1/4))√(1-4y²)+1-4y²=4y²±8y√(3y²-1/4)+4(3y²-1/4),±8y√(3y²-1/4)+4-48y²-2√(2-28y²±16y√(3y²-1/4)√(1-4y²)=0,64y²(3y²-1/4)±16y√(3y²-1/4)(4-48y²)+(4-48y²)²=4(2-28y²±16y√(3y ²-1/4)) 2496y⁴-288y²+8=±768y³√(3y²-1/ 4) 2496²y⁸+288²y⁴+8²-2*2496*288y⁶+ 2*2496*8y⁴-2*288*8y²=768²y⁶(3y²-1/4), (312²-96²*3)y⁸+(-624*36+96²/4)y⁶+(1296 +624)y⁴-72y²+1=0,y²=a,≥0,aє[1/12;1/4]144*(26²-64*3)a⁴+144(-52*3+16)a³+1920a²-72a+1=0,144*16*157a⁴-144*140a³+1920a²-72a+1=0, (а²-35/8/157а+0,00226..)² 7,28*10^-5а-4,29*10^-6=0,\>/> минимум а=35/8/157/2=35/16/157,-1,01*10^-6<0 поэтому есть 2 корня, а=0,014,а=0,010 у=±0,12..>0,289,<-0,289,у=±0,101>0,288,<-0,289 0/
169-12=157
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{logx+(logx+8logy)/(log²x+log²y)=2,logy+ (8logx-logy)/(log²x+log²y)=0;{log³x+logx log²y+logx+8logy=2log²x+2log²y,log²xlogy-log³y+8logx-logy=0;{log²y(logx-2)+8logy +log³x-2log²x+logx=0,log²xlogy-log³y+8logx-logy=0;{logy=(-8±√(64-4(logx-2)(log³x 2log²x+logx)))/2/(logx-2),[log²x(-8+√(-4lo g⁴x+16log³x-20log²x+8logx+64)/(2logx-4) -(-8+√(-4log⁴x+16log³x-20log²x+8logx+64)³/(2logx-4)³+8logx-(-8+√(-4log⁴x+16log³ x-20log²x+8logx+64)/(2logx-4)=0; log²x(-8-√(-4log⁴x+16log³x-20log²x+8logx+64)/(2logx-4)-(-8-√(-4log⁴x+16log³x-20lo g²x+8logx+64)/(2logx-4)+8logx-(-8-√(-4lo g⁴x+16log³x-20log²x+8logx+64)/(2logx-4) =0;(log²x-2logx+1/2)²-16,25≤0,(log²x-2lo gx-√16,25+0,5)(log²x-2logx+0,5+√16,25) ≤0,(logx-(2+√(4-4(-√16,25+0,5)))/2)(logx (2-√(4√16,25+2))/2)≤0,+*-*+>logx logxє [1-√(√16,25+0,5);1+√(√16,25+0,5)],[logx=0,logx=1,logx=2,logx=3,12,logxє [1-√(√16,25+0,5);1+√(√16,25+0,5)];[logx=0,=1,=2,=3,12;[x=1,x=10,x=100,x=1 0³,¹²;[y=1,y=10000;y=1,y=10⁸,y=10^(-1/4), y=10^(-25/7+√81678,7456/224),y=10^(-25/7-√81678,224).
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