1/k+k=5

(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

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)^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

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

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

100.01

or 100 1/100