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t=x²≥0
t²-5t-36=0
Δ=25+4*36=25+144=169
t=1/2*(5-13)=-8∉D v t=1/2*(5+13)=9
x²=9
x=-3 v x=3
h)
(x-1)(x+1)(2x-3)-(x+1)=0
(x+1)*[(x-1)(2x-3)-1]=0
(x+1)(2x²-3x-2x+3-1)=0
(x+1)(2x²-5x+2)=0
(x+1)(2x²-4x-x+2)=0
(x+1)*[2x(x-2)-(x-2)]=0
(x+1)(x-2)(2x-1)=0
x+1=0 v x-2=0 v 2x-1=0
Odp. x∈{-1, 1/2, 2}