Question

x la a doua+6x+8=6+4radical din 3
(3x+4)la a doua=(2x+1)la a doua
x la adoua+(x+2)la a doua+(x+4)la a doua=4(3x+8) va roggg mult fac orice numai sa-mi rezolvati si mie macar una.......ms mult...e de la lectia cu ecuatii de forma x la a doua=a

Answer 1

(3x+4)²=(2x+1)²
9x²+24x+16=4x²+4x+1
9x²-4x²+24x-4x+16-1=0
5x²+20x+15=0     :5
x²+4x+3=0
Δ=16-12=4
x1=(-4+2)/2=-1
x2=(-4-2)/2=-3
x∈{-1,-3}

x²+(x+2)²+(x+4)²=4(3x+8)
x²+x²+4x+4+x²+8x+16=12x+32
3x²+12x-12x+20-32=0
3x²-12=0
3x²=12
x²=12/3
x²=4
x1=√4=2
x2=-√4=-2
x∈{-2,2}