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प्रश्न
`a/((ax-1))+b/((bx-1))=(a+b),x≠1/a,1/b`
उत्तर
`a/((ax-1))+b/((bx-1))=(a+b) `
⇒`[a/((ax-1))-b]+[b/((bx-1))-a]=0`
⇒`(a-b(ax-1))/(ax-1)+(b-a(bx-1))/(bx-1)=0`
⇒`(a-abx+b)/(ax-1+a)-(abx+b)/(bx-1)=0`
⇒`(a-abx+b)[1/(ax-1)+1/(bx-1)]=0`
⇒`(a-abx+b)[((bx-1)+(ax-1))/(((ax-1))(bx-1))]=0`
⇒`(a-abx+b)[[(a+b)x-2)/((ax-1)(bx-1))]=0`
⇒`(a-abx+b)[(a+b)x-2]=0`
⇒`a-abx+b=0 or (a+b)x-2=0`
⇒`x=(a+b)/(ab) or x=2/(a+b)`
Hence, the roots of the equation are `(a+b)/(ab)` and` 2/(a+b)`
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