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=> x2 – 2x + 1 = 0 => (x-1)2 = 0 => x = 1 (x7 + x5 + x4 + x2)/x = (1 + 1 + 1 + 1)/1 = 4 Alternate method: If we put the value of x = 1 in the equation x2 – 2x + 1 = 0 So, LHS = RHS By putting the value of x = 1 in (x7 + x5 + x4 + x2)/x => (x7 + x5 + x4 + x2)/x => (1 + 1 + 1 + 1)/1 = 4
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2....
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