更新1:
可以加上一點中文解釋嗎@@ 有一點看不懂根號那邊
更新2:
就是大於等於2根號那
此圖跪求詳解 著急!!!! 求各位大大了?
回答 (2)
2019-07-22 10:20 am
Sol
a>0，b>0
(a-b)^2>=0
a^2-2ab+b^2>=0
a^2+b^2>=2ab
(a^2+b^2)/2>=ab
1
x>-1
x+1>0
x+9/(x+1)
(x+1)+9/(x+1)－1
=2*[(x+1)+9/(x+1)=/2－1
=2*[√(x+1)]^2+[3/√(x+1)]^2－1
>=2*3－1
=5
√(x+1)=3/√(x+1)
x+1=3
x=2
2
0<x<1
1/x>1
x+1/x>0
｜x+1/x｜=x+1/x
x－1/x<0
｜x－1/x｜=1/x－x
√(x^2+1/x^2+2)+√(x^2+1/x^2－2)
=√(x+1/x)^2+√(x－1/x)^2
=｜x+1/x｜+｜x－1/x)｜
=(x+1/x)+(1/x－x)
=2/x=16
x=1/8
a>0，b>0
(a-b)^2>=0
a^2-2ab+b^2>=0
a^2+b^2>=2ab
(a^2+b^2)/2>=ab
1
x>-1
x+1>0
x+9/(x+1)
(x+1)+9/(x+1)－1
=2*[(x+1)+9/(x+1)=/2－1
=2*[√(x+1)]^2+[3/√(x+1)]^2－1
>=2*3－1
=5
√(x+1)=3/√(x+1)
x+1=3
x=2
2
0<x<1
1/x>1
x+1/x>0
｜x+1/x｜=x+1/x
x－1/x<0
｜x－1/x｜=1/x－x
√(x^2+1/x^2+2)+√(x^2+1/x^2－2)
=√(x+1/x)^2+√(x－1/x)^2
=｜x+1/x｜+｜x－1/x)｜
=(x+1/x)+(1/x－x)
=2/x=16
x=1/8
2019-07-22 10:05 am
⒈
x + 9 / (x + 1)
= (x + 1) + 9 / (x + 1) - 1
(x + 1) + 9 / (x + 1) - 1 ≥ 2√{(x + 1)[9 / (x + 1)]} - 1
2√{(x + 1)[9 / (x + 1)]} - 1
= 2 · 3 - 1
= 5
此時 x + 1 = 9 / (x + 1)
x = 2
最小值 5，此時 x = 2
⒉
0 < x < 1
1 / x - x > 0
√(x² + 1 / x² + 2) + √(x² + 1 / x² - 2)
=√[(x + 1 / x)²] + √[(x - 1 / x)²]
= x + 1 / x + 1 / x - x
= 2 / x = 16
x = 1 / 8 #
x + 9 / (x + 1)
= (x + 1) + 9 / (x + 1) - 1
(x + 1) + 9 / (x + 1) - 1 ≥ 2√{(x + 1)[9 / (x + 1)]} - 1
2√{(x + 1)[9 / (x + 1)]} - 1
= 2 · 3 - 1
= 5
此時 x + 1 = 9 / (x + 1)
x = 2
最小值 5，此時 x = 2
⒉
0 < x < 1
1 / x - x > 0
√(x² + 1 / x² + 2) + √(x² + 1 / x² - 2)
=√[(x + 1 / x)²] + √[(x - 1 / x)²]
= x + 1 / x + 1 / x - x
= 2 / x = 16
x = 1 / 8 #
收錄日期: 2021-04-30 22:49:01
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20190722010511AAtzE8i