(2K)^2+[(根號3+1)K]^2-(根號2 K)^2 /2x2kx(根號3+1)k = 根號3/2 怎麼算的呢 請幫忙我解答謝謝?
回答 (2)
2016-04-04 5:09 am
先化簡分子 (2K)^2+[(根號3+1)K]^2-(根號2 K)^2= 4k²+(√3+1)k²-2k²=(6+2√3)k²
再來是分母 2x2kx(根號3+1)k=4kx(√3+1)k=(4√3+4)k²
k²消掉,就變成6+2√3/4√3+4
分子分母同乘以4√3-4
就變成(6+2√3)(4√3-4)/(4√3+4)(4√3-4)
化簡得16√3/32,也就是√3/2=根號3/2咯~
再來是分母 2x2kx(根號3+1)k=4kx(√3+1)k=(4√3+4)k²
k²消掉,就變成6+2√3/4√3+4
分子分母同乘以4√3-4
就變成(6+2√3)(4√3-4)/(4√3+4)(4√3-4)
化簡得16√3/32,也就是√3/2=根號3/2咯~
2016-04-03 6:19 pm
Please refer to the following link:
http://i707.photobucket.com/albums/ww74/stevieg90/ScreenHunter_994%20Apr.%2004%2001.16_zpszs6zuo6a.jpg
http://i707.photobucket.com/albums/ww74/stevieg90/ScreenHunter_994%20Apr.%2004%2001.16_zpszs6zuo6a.jpg
收錄日期: 2021-04-22 22:19:24
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https://hk.answers.yahoo.com/question/index?qid=20160403071103AADicAC