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-(3-4x)^2 + 3-4x+3=
2(-x^2 +x+3)/(-x^2 +x+3)^2 +1=
 rezolvati aceste exercitii 


Răspuns :

-(3-4x)² + 3-4x+3=-(9-24x+16x²)+6-4x=
-9+24x-16x²+6-4x=
-16x²+20x-3=0      (-1)
16x²-20x+3=0
Δ=b²-4ac=16²-4·16·3=256-192
Δ=64
x₁=-(-20)-√64)/2·16=(20-8)/32=12/32=3/8
x₂=(20+8)32=28/32=7/8

2(-x² +x+3)/(-x² +x+3)² +1=
[-2x
²+2x+6]/(x⁴-x³-3x²-x³+x²+3x-3x²+3x+9)=-1
(-2x²+2x+6)/(x⁴-2x³-5x²+6x+9)=-1
-2x²+2x+6=-(x⁴-2x³-5x²+6x+9)
-2x²+2x+6=-x⁴+2x³+5x²-6x-9
x⁴-2x³-2x²-5x²+2x+6x+6+9=0
x⁴-2x³-7x²+8x+15=0