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प्रश्न
`1/(x-2)+2/(x-1)=6/x,≠0,1,2`
उत्तर
`1/(x-2)+2/(x-1)=6/x`
⇒`((x-1)+2(x-2))/((x-1)(x-2))=6/x`
⇒ `(3x-5)/(x^2-3x+2)=6/x`
⇒`3x^2-5x=6x^2-18x+12` [On cross multiplying]
⇒`3x^2-13+12=0`
⇒`3x^2-(9+4)x+12=0`
⇒`3x^2-9x-4x+12=0`
⇒`3x(x-3)-4(x-3)=0`
⇒`(3x-4)(x-3)=0`
⇒`3x-4=0 or x-3=0`
⇒`x=4/3 or x=3`
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