Question

bracket x minus one upon X bracket close square minus 6 into bracket X + 1 upon X bracket close + 12 = 0​

Answer 1

the value of x =1 or 1/7..

Answer 2

Given :

$\left ( \frac{x-1}{x} \right )^2-\frac{6(x+1)}{x}+12=0$

Explanation:

$\left ( \frac{x-1}{x} \right )^2-\frac{6(x+1)}{x}+12=0\\\\using\\\\(a-b)^2= a^2+b^2-2ab\\\\\frac{x^2+1-2x}{x}-\frac{6x+6}{x}+12=0\\\\\text{taking lcm}\\\\\frac{x^2+1-2x-6x^2-6x+12x^2}{x^2}=0\\\\\text{cross multiplying}\\\\7x^2-8x+1=0\\\\7x^2-7x-x+1=0\\\\7x(x-1)-1(x-1)=0\\\\x-1=0 , 7x-1=0\\\\x = 1 , x = \frac{1}{7}$

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