Odpowiedź :
(2x+1)÷x=3x D:x∈R/{0}
(2x+1-3x²)÷x=0
(-3x²+2x+1)÷x=0
Δ=4+12=16
√Δ=4
x₁=(-2+4)÷-6=-1/3
x₂=(-2-4)÷-6=1
(2x+1-3x²)÷x=0
(-3x²+2x+1)÷x=0
Δ=4+12=16
√Δ=4
x₁=(-2+4)÷-6=-1/3
x₂=(-2-4)÷-6=1
2x+1/x=3x /×
2x+1=3x²
-3x²+2x+1+0
Δ=b2-4ac
Δ=4+12=16
√Δ=4
x1=-b+√Δ/2a
x1=-2+4/-6=-1/3
x2=-b-√Δ/2a
x2=-2-4/-6=1
2x+1=3x²
-3x²+2x+1+0
Δ=b2-4ac
Δ=4+12=16
√Δ=4
x1=-b+√Δ/2a
x1=-2+4/-6=-1/3
x2=-b-√Δ/2a
x2=-2-4/-6=1