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प्रश्न
`(x-4)/(x-5)+(x-6)/(x-7)=31/3,x≠5,7`
उत्तर
`(x-4)/(x-5)+(x-6)/(x-7)=3 1/3,x≠5,7`
⇒` ((x-4)(x-7)+(x-5)(x-6))/(x^2-12x+35)=10/3`
⇒`(x^2-11x+28+x^2-11x+30)/(x^2-12x+35)=10/3`
⇒`(2x^2-22x+58)/(x^2-12x+35)=10/3`
⇒` (x^2-22x+29)/(x^2-12x+35)=5/3`
⇒`(x^2-11x+29)/(x^2-12x+35)=5/3`
⇒`3x^2-33x+87=5x^2-60x+175`
⇒`2x^2-27x+88=0`
⇒`2x^2-16x11x+88=0`
⇒`2x(x-8)-11(x-8)=0`
⇒`(x-8) (2x-11)=0`
⇒`x-8=0 or 2x-11=0`
⇒`x=8 or x=11/2`
Hence, 8 and `11/2` are the roots of the given equation.
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