Question

Решите иррациональное неравенство

Answer 1

Пошаговое объяснение:

$\dfrac{\sqrt{2-x} -3+2x}{3-2x} <0\\\\\sqrt{2-x} -3+2x = 0\\\\\sqrt{2-x} =3-2x\\\\2-x=9-12x+4x^2\\\\4x^2-11x+7=0\\\\D=121-112=9\\\\x_1=1\\\\x_2=1,75\\\\3-2x=0\\\\x_3=1,5\\\\-----(1)---+--(1,5)-------(1,75)----[2]---\\\\x \in (-\infty;1)U(1,5;2]$

Answer 2

x∈(-∞;1)∪(1,5;2]

Пошаговое объяснение:

$\frac{\sqrt{2-x} }{3-2x} <1\\\left \{ {{2-x\geq0 } \atop {3-2x\neq 0}} \right. =$

x∈(-∞;1,5)∪(1,5;2]

$\frac{\sqrt{2-x}-3+2x }{3-2x} <0\\\sqrt{2-x}-3+2x=0\\\left \{ {{2-x\geq0 } \atop {3-2x\geq 0}} \right. =x\leq 1,5\\2-x=9-12x+4x^2<=4x^2-11x+7=0\\a+b+c=0=x_1=1;x_2=\frac{7}{4} =x=1\\-(-)-(1)-(+)-(1,5)-(-)-$

+ОДЗ