Question

question is in attachment​

Answer 1

$\boxed{\underline{\textsf{Solution :}}}$

$\textsf{Now,}\:\small{\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}}$

$=\small{\frac{(x-1)-(x-2)}{(x-1)(x-2)}+\frac{(x-2)-(x-3)}{(x-2)(x-3)}+\frac{(x-3)-(x-4)}{(x-3)(x-4)}}$

$=\small{\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}}$

$=-\frac{1}{x-1}+\frac{1}{x-4}$

$=\frac{-(x-4)+(x-1)}{(x-1)(x-4)}$

$=\frac{-x+4+x-1}{(x-1)(x-4)}$

$=\frac{3}{(x-1)(x-4)}$

$\textsf{Given that,}\:\frac{3}{(x-1)(x-4)}=\frac{1}{6}$

$\to (x-1)(x-4)=3*6$

$\to x^{2}-5x+4=18$

$\to x^{2}-5x-14=0$

$\to x^{2}-7x+2x-14=0$

$\to x (x-7)+2 (x-7)=0$

$\to (x-7)(x+2)=0$

$\textsf{Either, x - 7 = 0 or, x + 2 = 0}$

$\textbf{i.e., x = 7 , x = - 3}$

$\textbf{which is the required solution.}$

Answer 2

Evaluate the functions .

Go on till you reach a quadratic equation .

EXPLANATION :

A quadratic equation can be solved by middle split method .

We will split the middle part and then we will factorise .

After that we can apply the zero product rule and find the value of x .

SOLUTION :

$\frac{1}{(x-1)(x-2)}+\frac{1}{(x-2)(x-3)}+\frac{1}{(x-3)(x-4)}=\frac{1}{6}\\\implies \frac{(x-1)-(x-2)}{(x-1)(x-2)}+\frac{(x-2)-(x-3)}{(x-2)(x-3)}+\frac{(x-3)-(x-4)}{(x-3)(x-4)}=\frac{1}{6}\\\implies\frac{1}{x-2}-\frac{1}{x-1}+\frac{1}{x-3}-\frac{1}{x-2}+\frac{1}{x-4}-\frac{1}{x-3}=\frac{1}{6}\\\implies\frac{1}{x-4}-\frac{1}{x-1}=\frac{1}{6}\\\implies \frac{x+4-x-1}{(x-1)(x-4)}=\frac{1}6}\\\implies \frac{3}{(x-1)(x-4)}=\frac{1}{6}$

Cross multiply and we will get :

$(x-1)(x-4)=6\times3\\\\\implies x^2-4x-x+4=18\\\\\implies x^2-5x+4=18\\\\\implies x^2-5x-14=0\\\\\implies x^2-7x+2x-14=0\\\\\implies x(x-7)+2(x-7)\\\\\implies (x+2)(x-7)=0$

Either x is 7 or x is - 2 .