a)
(5-x)(2x+3)/5x+4*(-1)= 10x+15-2x²-3x/5x-4= -2x²+7x+15/5x-4
5x-4≠0 5x≠4 x≠⅘
Obliczamy deltę
Δ=b²-4ac= 49+120= 169
√169= 13
x₁= -b-√Δ/2a= -7-13/4= -20/4= -5
x₂= -b+√Δ/2a= -7+13/4= 6/4= 1½
x∈{-5,1½} z wyłączeniem ⅘
Nie jestem pewny końcówki sorry.
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