pigor: ..., no to może np. tak :
a)
2*5x−5x−1< 9 ⇔ 5
x−1(2*5−1)< 9 /:9 ⇔ 5
x−1< 1 ⇔
⇔ 5
x−1< 5
0 ⇔ x−1< 0 ⇔
x<1 ⇔
x∊(−∞;−1) . ...

−−−−−−−−−−−−−−−−−−−
a teraz znikam z forum
pigor: ..., no to dalej np. tak :
b)
25x−1−(14)2x > 0 ⇔ 2
5x−1 > 2
−2*2x ⇔ 5x−1 > −4x ⇔
⇔ 9x > 1 /:9 ⇔
x > 19 ⇔
x∊(19;+∞) .
−−−−−−−−−−−−−−−−−−−−−
c)
(13)x−4−9*3x < 0 ⇔ 3
−x+4 < 3
2+x ⇔
⇔ −x+4 < 2+x ⇔ 2 < 2x ⇔
x>1 ⇔
x∊(1;+∞) ,
−−−−−−−−−−−−−−−−−−−−−−
d)
8x+1−(14)2x−3 < 0 ⇔ 2
3x+3< 2
−4x+6 ⇔
⇔ 3x+3< −4x+6 ⇔ 7x< 3 ⇔
x < 37 ⇔
x∊(−∞;37} . ...