Answer:
Step-by-step explanation:
[tex]4\sqrt{3-\frac{1}{x} } -\sqrt{\frac{x}{3x-1} } =3\\4\sqrt{\frac{3x-1}{x} } -\sqrt{\frac{x}{3x-1} } =3\\\\put ~\sqrt{\frac{3x-1}{x}} =a\\4a-\frac{1}{a} =3\\4a^2-1-3a=0\\4a^2-4a+1a-1=0\\4a(a-1)+1(a-1)=0\\(a-1)(4a+1)=0\\a=1,-\frac{1}{4}\\[/tex]
[tex]\sqrt{\frac{3x-1}{x} } =1\\3x-1=x\\3x-x=1\\2x=1\\x=\frac{1}{2}[/tex]
when a=-1/4
[tex]\sqrt{\frac{3x-1}{x} } =-\frac{1}{4} \\[/tex]
[tex]\frac{3x-1}{x} =\frac{1}{16} \\48x-16=x\\47x=16\\x=\frac{16}{47}[/tex]
[tex]\frac{(1-2x)^2 \times (x^2-9)}{-x^2-x+6} \geq 0,\\both ~numerator ~and~denominator ~are~of~same~sign.\\\frac{(1-2x)^2(x^2-9)}{-x^2-3x+2x+6} \geq 0\\\frac{(1-2x)^2(x+3)(x-3)}{-x(x+3)+2(x+3)} \geq 0\\\frac{(1-2x)^2(x+3)(x-3)}{(x+3)(-x+2)} \geq 0\\(1-2x)^2\geq 0~(always)\\x\neq -3\\case.~1.\\both~numerator~and~denominator >0\\\frac{x-3}{-x+2} \geq 0\\x-3\geq 0\\x\geq 3\\-x+2> 0\\-x>-2\\x<2\\so~ solution~ is~ (-\infty,-3)U(-3,2)U[3,\infty)[/tex]
case 2.
both numerator and denominator are negative.
[tex]x-3\leq 0\\x\leq 3\\-x+2\leq 0\\-x\leq -2\\x\geq 2\\solution~is~(-\infty,-3)U(-3,3]U[2,\infty)[/tex]
