Rozwiąż nierówność:
\sqrt{1+x^{2}}>-x //^{2}
1+x^{2}>(-x^{2})
1+x^{2}>x^{2}
1>x^{2}-x^{2}
1>0
x należy do R
V(3-2x-x^2) > x+1 I^2
3-2x-x^2 > (x+1)^2
3-2x-x^2 > x^2 +2x+1
-2x^2 -4x +2 > 0 I:2
-x^2 -2 +1 > 0
D =b^2 -4ac
D =(-2)^2 -4*(-1)*1
D = 8
VD =2V2
X1 =(2 -2V2)/(-2) =2(1-V2)/(-2) =-(1-V2)
X1 =V2+1
========
X2 =(2+2V2)/(-2) =2(1+V2)/(-2)
X2 =-V2-1
=========
a < 0nie wiem czy dobrze
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\sqrt{1+x^{2}}>-x //^{2}
1+x^{2}>(-x^{2})
1+x^{2}>x^{2}
1>x^{2}-x^{2}
1>0
x należy do R
V(3-2x-x^2) > x+1 I^2
3-2x-x^2 > (x+1)^2
3-2x-x^2 > x^2 +2x+1
-2x^2 -4x +2 > 0 I:2
-x^2 -2 +1 > 0
D =b^2 -4ac
D =(-2)^2 -4*(-1)*1
D = 8
VD =2V2
X1 =(2 -2V2)/(-2) =2(1-V2)/(-2) =-(1-V2)
X1 =V2+1
========
X2 =(2+2V2)/(-2) =2(1+V2)/(-2)
X2 =-V2-1
=========
a < 0
nie wiem czy dobrze