標題:
續二次方程(11-14題)
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發問:
11.解1+ √x+6 = 2/√x+6 12.解 2/x-3 +3 =7/√x-3 13.解2^2x - 3(2^x+2) + 32 =0 14.解 9^x + 2(3^x) - 3 =0
最佳解答:
11. 設 u = √(x + 6) 1 + √(x + 6) = 2/√(x + 6) 1 + u = 2/u u^2 + u - 2 = 0 (u - 1)(u + 2) = 0 u = 1 或 u = -2 √(x + 6) = 1 或 √(x + 6) = -2(不合) x + 6 = 1 x = -5 ===== 12. 設 u = 1/√(x - 3) [2/(x - 3)] + 3 = 7/√(x - 3) 2u^2 + 3 = 7u 2u^2 - 7u + 3 = 0 (2u - 1)(u - 3) = 0 u = 1/2 或 u = 3 1/√(x - 3) = 1/2 或 1/√(x - 3) =3 √(x - 3) = 2 或 √(x - 3) = 1/3 x - 3 = 4 或 x - 3 = 1/9 x = 7 或 x = -28/9 ===== 13. 設 u = 2^x 2^2x - 3[2^(x+2)] +32 = 0 (2^x)^2 - 3[(2^2)*(2^x)] + 32 = 0 (2^x)^2 - 12(2^x) + 32 = 0 u^2 - 12u + 32 = 0 (u - 4)(u - 8) = 0 u = 4 或 u = 8 2^x = 2^2 或 2^x = 2^3 x = 2 或 x = 3 ===== 14. 設 u = 3^x 9^x + 2(3^x) - 3 =0 (3^x)^2 + 2(3^x) - 3 = 0 u^2 + 2u - 3 = 0 (u - 1)(u + 3) = 0 u = 1 或 u = -3 3^x = 3^0 或 3^x = -3(不合) x = 0
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