1/(x-1)(x-2)+1/(x-2)(x-3)+1/(x-3)(x-4)=1/6
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Poulami Dasgupta 6 years, 9 months ago
[{tex}1\over(x-1)(x-2){/tex}+{tex}1\over(x-2)(x-3){/tex}]+{tex}1\over(x-3)(x-4){/tex}={tex}1\over6{/tex}
=>{tex}(x-3)+(x-1)\over(x-1)(x-2)(x-3){/tex}+{tex}1\over (x-3)(x-4){/tex}={tex}1\over6{/tex}
=>{tex}2x-4\over(x-1)(x-2)(x-3){/tex}+{tex}1\over(x-3)(x-4){/tex}={tex}1\over6{/tex}
=>{tex}2(x-2)\over(x-1)(x-2)(x-3){/tex}+{tex}1\over(x-3)(x-4){/tex}={tex}1\over6{/tex}
=>{tex}2\over(x-1)(x-3){/tex}+{tex}1\over(x-3)(x-4){/tex}={tex}1\over6{/tex}
=>{tex}2(x-4)+(x-1)\over(x-1)(x-3)(x-4){/tex}={tex}1\over6{/tex}
=>{tex}2x-8+x-1\over(x-1)(x-3)(x-4){/tex}={tex}1\over6{/tex}
=>{tex}3x-9\over(x-1)(x-3)(x-4){/tex}={tex}1\over6{/tex}
=>{tex}3(x-3)\over(x-1)(x-3)(x-4){/tex}={tex}1\over6{/tex}
=>{tex}3\over(x-1)(x-4){/tex}={tex}1\over6{/tex}
=>{tex}(x-1)(x-4)=18{/tex}
=>{tex}x\ ^2-4x-x+4=18{/tex}
=>{tex}x\ ^2-5x-14=0{/tex}
=>{tex}x\ ^2-7x+2x-14=0{/tex}
=>{tex}x(x-7)+2(x-7)=0{/tex}
=>{tex}(x+2)(x-7)=0{/tex}
=>{tex}either ,x+2=0 or, x-7=0{/tex}
{tex}x=-2 or7{/tex}
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