1/k+k=5
回答 (6)
2007-11-17 3:30 am
✔ 最佳答案
(1/k)^2+k^2=(1/k)^2+2(1/k)(k)+k^2-2(1/k)(k)
=(1/k+k)^2-2
=25-2
=23
參考: my mathematical knowledge
2007-11-17 10:29 pm
1/k+k=5
1/k+(k^2/k)=5
(1+k^2)/k=5
1+k^2=5k
k^2=5k-1
k^2-5k+1=0
k={-(-5)+or- [surd(-5)^2-4(1)(1)]}/2(1)
k=[5+(surd21)]/2 or [5-(surd21)]/2
Hence,put k into 1/k^2+k^2
1/k^2+k^2= 23
1/k+(k^2/k)=5
(1+k^2)/k=5
1+k^2=5k
k^2=5k-1
k^2-5k+1=0
k={-(-5)+or- [surd(-5)^2-4(1)(1)]}/2(1)
k=[5+(surd21)]/2 or [5-(surd21)]/2
Hence,put k into 1/k^2+k^2
1/k^2+k^2= 23
參考: Me
2007-11-17 5:47 am
(1/k)^2+k^2
=((1/k)^2+2(1/k)(k)+k^2)-2(1/k)(k)
=(1/k+k)^2-2 -->(1/k+k)^2 = (1/k)^2+2(1/k)(k)+k^2 , (1/k)(k) = 1
=5^2-2
=25-2
=23
=((1/k)^2+2(1/k)(k)+k^2)-2(1/k)(k)
=(1/k+k)^2-2 -->(1/k+k)^2 = (1/k)^2+2(1/k)(k)+k^2 , (1/k)(k) = 1
=5^2-2
=25-2
=23
2007-11-17 3:36 am
Method 1
1/k+k=5
k^2-5k+1=0
By using the quadratic formula, k=4.79 or 0.209 (Corr. to 3 sig.fig)
Sub. k into the lower eqn, we have 1/k^2+k^2=23
Method 2
Since 1/k+k=5
Then, (1/k+k)^2=5^2=25
1/k^2+2+k^2=25
Therefore1/k^2+k^2=23
1/k+k=5
k^2-5k+1=0
By using the quadratic formula, k=4.79 or 0.209 (Corr. to 3 sig.fig)
Sub. k into the lower eqn, we have 1/k^2+k^2=23
Method 2
Since 1/k+k=5
Then, (1/k+k)^2=5^2=25
1/k^2+2+k^2=25
Therefore1/k^2+k^2=23
參考: 自己
2007-11-17 3:34 am
1/k+k=5
將成條式乘k
即係變左1+k^2=5k
之後調位變左k^2-5k+1=0
解2次方程
要用-b正負b^2-4ac的開方/2a果條式
答案係5正負21的開方/2
將成條式乘k
即係變左1+k^2=5k
之後調位變左k^2-5k+1=0
解2次方程
要用-b正負b^2-4ac的開方/2a果條式
答案係5正負21的開方/2
2007-11-17 3:26 am
100.01
or 100 1/100
or 100 1/100
收錄日期: 2021-05-02 21:01:39
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