F.5 APPLICATIONS OF DIFFERENTI

2011-10-23 7:13 am

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圖中的第10題小弟唔太肯定點做

請師兄教教小弟!

THZ

更新1:

師兄我想問一下 x/√(x^2 + 4) + (x - 10)/√[(10 - x)^2 + 9] = 0 x^2/(x^2 + 4) = (x - 10)^2/[(10 - x)^2 + 9] <--依度點解可以將下面個1/2次變番1次上面1次變2次既?.? 依個咩方法黎???

回答 (1)

myisland8132

2011-10-23 8:08 am

✔ 最佳答案

Let BP = x and CP = 10 - xAP = √(x^2 + 4)DP = √[(10 - x)^2 + 9]AP + DP = L = √(x^2 + 4) + √[(10 - x)^2 + 9]dL/dx= x/√(x^2 + 4) + (x - 10)/√[(10 - x)^2 + 9]Let dL/dx = 0x/√(x^2 + 4) + (x - 10)/√[(10 - x)^2 + 9] = 0x^2/(x^2 + 4) = (x - 10)^2/[(10 - x)^2 + 9](x - 10)^2 (x^2 + 4) = x^2[(10 - x)^2 + 9](x^2 - 20x + 100)(x^2 + 4) = x^4 - 20x^3 + 109x^2x^4 - 20x^3 + 104x^2 - 80x + 400 = x^4 - 20x^3 + 109x^2x^2 + 16x - 80 = 0(x + 20)(x - 4) = 0x = 4 or -20 (rejected)So, BP = 4m

2011-10-23 12:16:47 補充：
將兩邊平方就可以了

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