Question

15^4+x - 27^x*25^2x-1/ под корнем 6-x меньше или равно 0 

Answer 1

$frac{15^{4+x}-27^x*25^{2x-1}}{sqrt{6-x}} leq 0\ \ left { {{15^{4+x}-27^x*25^{2x-1} leq 0} atop {6-x > 0}} right. \ left { {{3^{4+x}*5^{4+x}-3^{3x}*5^{4x-2} leq 0} atop {6-x >0}} right.\ left { {{81*3^x*625*5^x-3^{3x}*5^{4x-2} leq 0} atop {x<6}} right. \ left { {{frac{15^x}{25}*(5^6*81-3^{2x}*5^{3x}) leq 0 } atop {x<6}} right. \ \ left { {{5^6*81-3^{2x}*5^{3x} leq 0} atop {x<6}} right. \ left { {{1125^x geq 1125^2} atop { x<6 }} right.$
и того $2 leq x<6$