можете решить B4, C2, C3???

Ответ
log(0,3) [x+1] / { log(0,3) 100 - log(0,3) 9} < 1
ОДЗ: [x+1]>0 => x>1
log(0,3) [x+1] < { log(0,3) 100 - log(0,3) 9}
log(0,3) [x+1] < log(0,3) 100/9
[x+1] > 100/9
9x + 9 > 100
9x > 91
x > 91/9

[V(7 + 4V3)]^x + [V(7 - 4V3)]^x = 14
__ 7 + 4V3 = 4+3 + 4V3 = 2^2 + 2*2*V3 + (V3)^2 = (2+V3)^2
__ 7 - 4V3 = 4+3 - 4V3 = 2^2 - 2*2*V3 + (V3)^2 = (2-V3)^2
=>
[V{(2+V3)^2}]^x + [V{(2-V3)^2}]^x = 14
[+-(2+V3)]^x + [+-(2-V3)]^x = 14
Дальше легко.