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14 Solve the equation 5x+1 = 3x Solution: 5x+1 = 3x Equations 2 −1 . 2 −1 ⇔ (x + 1) lg 5 = (x2 − 1) lg 3 ⇔ (x + 1)[lg 5 − (x − 1) lg 3] = 0 , thus x + 1 = 0 or lg 5 − (x − 1) lg 3 = 0 , which imply two solutions x = −1 , x = log3 15 . 15 Solve the equation x4 − 4x2 + 1 = 0 . Solution: Let y = x2 , then the equation becomes √ √ √ y 2 − 4y + 1 = 0 ⇔ (y − 2)2 = 3 ⇒ y = 2 ±√ 3 ⇒ x2 = 2 + 3 or x2 = 2 − 3 , √ of which implies the first of which implies x = ± 2 + 3 = ± 3+1 2 , the second √ √ √ √ √ √ 3+1 3+1 3−1 3−1 3−1 , − , , − .

Then x(x + 2) = z + 120 ⇔ z = (x − 10)(x + 12) . Since z is a prime number, then x − 10 = 1 , then x = 11, z = 23 . As a conclusion, there are two possibilities: x = 2, y = 59, z = 2 or x = 11, y = 2, z = 23 . 31 Solve the equation √ 2x2 − 7x + 1 − √ 2x2 − 7x + 1 + 2x2 − 9x + 4 : √ √ √ 2x2 − 7x + 1 + 2x2 − 9x + 4 = 2x − 3 (ii). (i)+(ii): 2x2 − 7x + 1 = x − 1 , taking square to obtain x2 − 5x = 0 ⇒ x(x − 5) = 0 ⇒ x = 0 or x = 5 . We can verify these two possible solutions via the original equation (i): x = 5 is indeed a root of (i), but x = 0 is a extraneous root Solution: Multiply both sides by √ √ 2x2 − 9x + 4 = 1 (i).

Since 2 and 5 are coprime, thus at least one of a3 b − ab3 = ab(a2 − b2 ), b3 c − bc3 = bc( = ab(a2 − b2 ), b3 c − bc3 = bc(b2 − c2 ), c3 a − ca3 = ca(c2 − a2 ) is divisible by 10. 108 Let a, b, c are positive integers and follow a geometric sequence, and b − a is a perfect square, log6 a + log6 b + log6 c = 6 , find the value of a + b + c . Solution: log6 a + log6 b + log6 c = 6 ⇒ log6 abc = 6 ⇒ abc = 66 . In addition, b2 = ac , then b = 62 = 36, ac = 362 . In order to make 36 − a a perfect square, a can only be 11,27,32,35, and a is a divisor of 362 , thus a = 27 , then c = 48 .