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Find the least integral value of f(x)=(x−1)7+3.(x−1)6+(x−1)5+1(x−1)5,∀x>1f(x)=\dfrac{(x-1)^7+3.(x-1)^6+(x-1)^5+1}{(x-1)^5} \quad \quad , \forall x>1f(x)=(x−1)5(x−1)7+3.(x−1)6+(x−1)5+1,∀x>1
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