Question

5х-4/6-1>2х+1/3. 3х+1/4-2х>2,5-3х-2/8

Answer 1

Пошаговое объяснение:

$\frac{5x-4}{6-1} >\frac{2x+1}{3}$

$3(5x-4)>5(2x+1)\\15x-12>10x+5\\15x-10x>5+12\\5x>17\\x>\frac{17}{5}$

Ответ :

x ∈ ( $\frac{17}{5}$; +∞)

$\frac{3x+1}{4-2x} >\frac{2,5-3x-2}{8}$

$\frac{3x+1}{4-2x} >\frac{2,5-3x-2}{8} , x\neq 2\\\frac{3x+1}{4-2x} -\frac{2,5-3x-2}{8}>0\\\frac{3x+1}{2(2-x)} -\frac{2,5-3x-2}{8}>0\\\\\frac{4(3x+1)-(2-x)*(2,5-3x-2)}{8(2-x)}>0\\\frac{12x+4-(1-6x-0,5x+3x^{2} )}{16-8x} >0\\\frac{12x+4-1+6,5x-3x^{2} }{16-8x} >0\\\frac{37x+6-6x^{2} }{2(16-8x)} >0\\\left \{ {{37x+6-6x^{2} >0} \atop {2(16-8x)>0}} \right. \\\left \{ {{37x+6-6x^{2}<0 } \atop {2(16-8x)<0}} \right.$

x ∈ $(\frac{37-\sqrt{1513} }{12} ,2)$

x ∈ $(\frac{37+\sqrt{1513} }{12} ,$+∞)

x ∈ $(\frac{37-\sqrt{1513} }{12} ,2)$ ∪ $(\frac{37+\sqrt{1513} }{12} ,$+∞)

Ответ:

x ∈ $(\frac{37-\sqrt{1513} }{12} ,2)$ ∪ $(\frac{37+\sqrt{1513} }{12} ,$+∞)