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l2x+6l=radical 8(modul de 2x+6)


Răspuns :

I2x+6I=√8
=>2Ix+3I=2√2
=>Ix+3I=√2
=> x+3= -√2 sau x+3=√2
x+3= -√2=>x=-√2-3
x+3=√2=> x=√2-3
x∈{(-√2-3),(√2-3)}