Lg(x²-2x-2)<1<br>D(y): x²-2x-2>0
D=(-2)²-4*1*(-2)=4+8=12
√D=√(4*3)=2√3
x=(2+2√3)/2=1+√3
x=1-√3
(x-(1+√3))(x-(1-√3))>0
///////////// ////////////
_______o_________o______
1-√3 1+√3
x∈(-∞;1-√3)U(1+√3;+∞)
lg(x²-2x-2)<1<br>lg(x²-2x-2)т.к. основание логафрима больше 1, то знак неравенства не меняется:
x²-2x-2<10<br>x²-2x-12<0<br>D=(-2)²-4*(-12)=4+48=52
√D=√(4*13)=2√13
x=(2+2√13)/2=1+√13
x=1-√13
(x-(1-√13))(x-(1+√13))<0<br>\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\ \\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
///////////////////////////////////////////////////////
________o_________o_________o_________o________
1-√13 1-√3 1+√3 1+√13
Ответ: x∈(1-√13;1-√3)U(1+√3;1+√13)