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Tricky. I plugged in x=7 and x=5.1, and it works.
Alguien podra decirme como resolver la siguiente ecuacion? (√x) +1 =0. Parece muy obvio, pero me gustaria saber su opinion. Gracias
[log5 + log(x - 5)]log(x - 5) = log2log(x - 5) = uu² + ulog5 - log2 = 0u = [-log5 ± √(log²5 + 4log2)]/2log²5 = (1 - log2)² = log²2 + 1 - 2log2u = [-log5 ± √(1 + log2)²]/2u = [-log5 ± (log20)]/2u = (log4)/2 ∨ u = -2/2u = log2 ∨ u = -1log(x - 5) = log2 => *x = 7*log(x - 5) = -1 => *x = 51/10*
won't work!
@swedishpsychopath8795 your brain?
=>> y=logm; y²+ylog5+log2=0 ;Let log5=a; log2 =log5/10=log5-log10=a-1; =>y²+ay+(a-1)=0;y=(-a±√{a² -4(a-1)}/2y={-a±(a-2)}/2;y=-1 or (-a+1); logm=-1=log1/10,=> m=1/10 ,>(x=5.1)logm=-a+1=-log5+log10=log10/5; m=2;(x=7)
Explain@@swedishpsychopath8795
Check with x= 5.1 or 7 @@swedishpsychopath8795
Tricky. I plugged in x=7 and x=5.1, and it works.
Alguien podra decirme como resolver la siguiente ecuacion? (√x) +1 =0. Parece muy obvio, pero me gustaria saber su opinion. Gracias
[log5 + log(x - 5)]log(x - 5) = log2
log(x - 5) = u
u² + ulog5 - log2 = 0
u = [-log5 ± √(log²5 + 4log2)]/2
log²5 = (1 - log2)² = log²2 + 1 - 2log2
u = [-log5 ± √(1 + log2)²]/2
u = [-log5 ± (log20)]/2
u = (log4)/2 ∨ u = -2/2
u = log2 ∨ u = -1
log(x - 5) = log2 => *x = 7*
log(x - 5) = -1 => *x = 51/10*
won't work!
@swedishpsychopath8795 your brain?
=>> y=logm;
y²+ylog5+log2=0 ;
Let log5=a; log2 =log5/10=log5-log10
=a-1; =>y²+ay+(a-1)=0;
y=(-a±√{a² -4(a-1)}/2
y={-a±(a-2)}/2;
y=-1 or (-a+1); logm=-1=log1/10,=> m=1/10 ,>(x=5.1)
logm=-a+1=-log5+log10
=log10/5; m=2;(x=7)
won't work!
Explain@@swedishpsychopath8795
Check with x= 5.1 or 7
@@swedishpsychopath8795