Rovnice vzorec
12/1-9xna druhou = 1-3x/1+3x + 1+3x/3x-1 vim, ze to je vzorec, ale nevim jak to udelat na te prave strane.
Odpovědi
Diskuze
12/(1-9x²) = (1-3x)/(1+3x) + (1+3x)/(3x-1)
12/(1-9x²) = (1-3x)/(3x+1) + (1+3x)/(3x-1)
12/(1-9x²) = ( (1-3x)*(3x-1) + (3x+1)*(1+3x) )/( (3x+1)*(3x-1) )
12/(1-9x²) = ( -(3x-1)*(3x-1) + (9x²+2*3x+1) )/(9x²-1)
12/(1-9x²) = ( -(9x²-2*3x+1) + (9x²+2*3x+1) )/(9x²-1)
12/(1-9x²) = (4*3x)/(9x²-1)
12/(1-9x²) = -(4*3x)/(1-9x²)
12 = -(4*3x)
12 = -12x
x=-1
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12/(1-9x²) = (1-3x)/(1+3x) + (1+3x)/(3x-1)
12/(1-9) = (1+3)/(1-3) + (1-3)/(-3-1)
12/(-8) = 4/(-2) + (-2)/(-4)
-3/2 = -2 + 1/2
-3 = -4 + 1
-3=-3; OK
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Dále musí platit:
(1-9x²) <> 0
-1/9 <> x²
√(-1/9) <> x²
x <> ±1/3i
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Pravá strana:
1-3x/1+3x + 1+3x/3x-1 =
1-3x/1+3x + 1+3x/-(-3x+1) =
1-3x/1+3x - 1+3x/(-3x+1) =
1-3x/1+3x - 1+3x/(1-3x) =
společný násobek je (1+3x).(1-3x)
[(1-3x)(1-3x) - (1+3x).(1+3x)]/(1+3x).(1-3x) =
[(1-3x)^2 - (1+3x)^2]/(1-9x^2) =
...
vynásobit obě strany rovnice (1-9x^2), tím se odstraní zlomky na obou stranách
12 = (1-3x)^2 - (1+3x)^2
napravo opět vzorec a^2 - b^2
12 = [(1-3x)+(1+3x)] . [(1-3x)-(1+3x)]
12 = [1-3x+1+3x] . [1-3x-1-3x]
12 = 2.(-6x)
x = -1