Question

3x-1supra 4=x+1supra2x ajuta.tima va rog

Answer 1

$\frac{3x-1}{4} = \frac{x+1}{2x} \\ 2x(3x-1)=4(x+1) \\ 6 x^{2} -2x=4x+4 \\ 6 x^{2} -6x-4=0 \\ 3 x^{2} -3x-2=0$
Δ=b²-4ac=(-3)²+4*3*2=9+24=33

x₁=$\frac{-b+ \sqrt{d} }{2a} = \frac{3+ \sqrt{33} }{6}$
x₂=$\frac{-b+ \sqrt{d} }{2a} = \frac{3- \sqrt{33} }{6}$
d=Δ

[tex] \frac{3x-5}{2} = \frac{2x+1}{x} \\ x(3x-5)=2(2x+1) \\ 3x^{2} -5x=4x+2 \\ 3x^{2} -9x-2=0 [/tex]
Δ=b²-4ac=9²+4*3*2=81+24=105

x₁=$\frac{-b+ \sqrt{d} }{2a}= \frac{9+ \sqrt{105} }{18}$
x₂=$\frac{-b- \sqrt{d} }{2a} = \frac{9 \sqrt{105} }{18}$