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1÷(x-1)(x-2)+1÷(x-2)(x-3)+1÷(x-3)(x-4)=1/6 solve the value of x

Answer» We have,{tex}\\frac{1}{{(x - 1)(x - 2)}} + \\frac{1}{{(x - 2)(x - 3)}}{/tex}{tex}+ \\frac{1}{{(x - 3)(x - 4)}} = \\frac{1}{6}{/tex}{tex}\\Rightarrow{/tex}{tex} (x - 3)(x - 4) + (x - 1)(x - 4) + (x - 1)(x - 2) = {/tex}{tex}\\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}[{tex}\\because{/tex} Multiplying both sides by (x -1)(x - 2)(x - 3)(x - 4)]{tex}\\Rightarrow{/tex}\xa0{tex}x^2 - 4x - 3x + 12 + x^2 - 4x - x + 4 + x^2 - 2x - x + 2 ={/tex}{tex}\\frac{1}{6}{/tex}{tex}[(x - 1)(x - 2)(x - 3)(x - 4)]{/tex}{tex}\\Rightarrow{/tex}{tex}3x^2 - 15x + 18 ={/tex}{tex}\\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}3(x^2 - 5x + 6) =\u200b\u200b\u200b\u200b\u200b\u200b\u200b{/tex}{tex}\\frac{1}{6}{/tex}{tex}(x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}18[x^2 - 3x - 2x + 6] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}18[x(x - 3) - 2(x - 3)] = (x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex}{tex}18(x - 3)(x - 2) = (x - 1)(x - 2)(x - 3)(x - 4){/tex}{tex}\\Rightarrow{/tex} 18 = (x - 1)(x - 4){tex}\\Rightarrow{/tex} 18 = x2 - 4x - 1x + 4{tex}\\Rightarrow{/tex} x2 - 5x + 4 - 18 = 0{tex}\\Rightarrow{/tex} x2 - 5x - 14 = 0In order to factorize x2 - 5x - 14, we have to find two numbers \'a\' and \'b\' such that.a + b = - 5 and ab = -14Clearly, -7 + 2 = -5 and (-7)(2) = -14{tex}\\therefore{/tex}\xa0a = -7 and b = 2Now,x2 - 5x - 14 = 0{tex}\\Rightarrow{/tex} x2 - 7x + 2x - 14 = 0{tex}\\Rightarrow{/tex} x(x - 7) + 2(x - 7) = 0{tex}\\Rightarrow{/tex} (x - 7)(x + 2) = 0{tex}\\Rightarrow{/tex} x - 7 = 0 or x + 2 = 0{tex}\\Rightarrow{/tex} x = 7 or x = -2


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