

Chứng minh rằng:
a)
`1/(sqrt(1) + sqrt(3)) + 1/(sqrt(5) + sqrt(7)) + 1/(sqrt(9)+sqrt(11)) + ... + 1/(sqrt(97)+sqrt(99)) > 9/4`
b)
`1/(sqrt(1) + sqrt(2)) + 1/(sqrt(3) + sqrt(4)) + 1/(sqrt(5)+sqrt(6)) + ... + 1/(sqrt(79)+sqrt(80)) > 4`
Hãy luôn nhớ cảm ơn và vote 5*
nếu câu trả lời hữu ích nhé!
`@` `\color{Silver}{\text{πk2}`
-----------------
`a)` Ta có:
`1/(\sqrt{1}+\sqrt{3})>1/(\sqrt{3}+\sqrt{5})`
`1/(\sqrt{5}+\sqrt{7})>1/(\sqrt{7}+\sqrt{9})`
`...`
`1/(\sqrt{97}+\sqrt{99})>1/(\sqrt{99}+\sqrt{101})`
`⇒2·(1/(\sqrt{1}+\sqrt{3})+1/(\sqrt{5}+\sqrt{7})+1/(\sqrt{9}+\sqrt{11})+...+1/(\sqrt{97}+\sqrt{99}))`
`>1/(\sqrt{1}+\sqrt{3})+1/(\sqrt{3}+\sqrt{5})+1/(\sqrt{5}+\sqrt{7})+1/(\sqrt{7}+\sqrt{9})+1/(\sqrt{9}+\sqrt{11})+...+1/(\sqrt{97}+\sqrt{99})+1/(\sqrt{99}+\sqrt{101})`
`=1/2·(2/(\sqrt{1}+\sqrt{3})+2/(\sqrt{3}+\sqrt{5})+2/(\sqrt{5}+\sqrt{7})+2/(\sqrt{7}+\sqrt{9})+2/(\sqrt{9}+\sqrt{11})+...+2/(\sqrt{97}+\sqrt{99})+2/(\sqrt{99}+\sqrt{101}))`
`=1/2·((3-1)/(\sqrt{3}+\sqrt{1})+(5-3)/(\sqrt{5}+\sqrt{3})+(7-5)/(\sqrt{7}+\sqrt{5})+(9-7)/(\sqrt{9}+\sqrt{7})+(11-9)/(\sqrt{11}+\sqrt{9})+...+(99-97)/(\sqrt{99}+\sqrt{97})+(101-99)/(\sqrt{101}+\sqrt{99}))`
`=1/2·[[(\sqrt{3}-\sqrt{1})(\sqrt{3}+\sqrt{1})]/(\sqrt{3}+\sqrt{1})+[(\sqrt{5}-\sqrt{3})(\sqrt{5}+\sqrt{3})]/(\sqrt{5}+\sqrt{3})+[(\sqrt{7}-\sqrt{5})(\sqrt{7}+\sqrt{5})]/(\sqrt{7}+\sqrt{5})+[(\sqrt{9}-\sqrt{7})(\sqrt{9}+\sqrt{7})]/(\sqrt{9}+\sqrt{7})+[(\sqrt{11}-\sqrt{9})(\sqrt{11}+\sqrt{9})]/(\sqrt{11}+\sqrt{9})+...+[(\sqrt{99}-\sqrt{97})(\sqrt{99}+\sqrt{97})]/(\sqrt{99}+\sqrt{97})+[(\sqrt{101}-\sqrt{99})(\sqrt{101}+\sqrt{99})]/(\sqrt{101}+\sqrt{99})]`
`=1/2·(\sqrt{3}-\sqrt{1}+\sqrt{5}-\sqrt{3}+\sqrt{7}-\sqrt{5}+\sqrt{9}-\sqrt{7}+\sqrt{11}-\sqrt{9}+...+\sqrt{99}-\sqrt{97}+\sqrt{101}-\sqrt{99})`
`=1/2·(\sqrt{101}-1)`
`>1/2·(\sqrt{100}-1)`
`=1/2·(10-1)`
`=1/2·9`
`=9/2`
`⇒2·(1/(\sqrt{1}+\sqrt{3})+1/(\sqrt{5}+\sqrt{7})+1/(\sqrt{9}+\sqrt{11})+...+1/(\sqrt{97}+\sqrt{99}))>9/2`
`⇒1/(\sqrt{1}+\sqrt{3})+1/(\sqrt{5}+\sqrt{7})+1/(\sqrt{9}+\sqrt{11})+...+1/(\sqrt{97}+\sqrt{99})>9/4`
-------------------------
Ta có:
`1/(\sqrt{1}+\sqrt{2})>1/(\sqrt{2}+\sqrt{3})`
`1/(\sqrt{3}+\sqrt{4})>1/(\sqrt{4}+\sqrt{5})`
`...`
`1/(\sqrt{79}+\sqrt{80})>1/(\sqrt{80}+\sqrt{81})`
`⇒2·(1/(\sqrt{1}+\sqrt{2})+1/(\sqrt{3}+\sqrt{4})+1/(\sqrt{5}+\sqrt{6})+...+1/(\sqrt{79}+\sqrt{80}))`
`>1/(\sqrt{1}+\sqrt{2})+1/(\sqrt{2}+\sqrt{3})+1/(\sqrt{3}+\sqrt{4})+1/(\sqrt{4}+\sqrt{5})+1/(\sqrt{5}+\sqrt{6})+...+1/(\sqrt{79}+\sqrt{80})+1/(\sqrt{80}+\sqrt{81})`
`=(2-1)/(\sqrt{2}+\sqrt{1})+(3-2)/(\sqrt{3}+\sqrt{2})+(4-3)/(\sqrt{4}+\sqrt{3})+(5-4)/(\sqrt{5}+\sqrt{4})+(6-5)/(\sqrt{6}+\sqrt{5})+...+(80-79)/(\sqrt{80}+\sqrt{79})+(81-80)/(\sqrt{81}+\sqrt{80})`
`=[(\sqrt{2}-\sqrt{1})(\sqrt{2}+\sqrt{1})]/(\sqrt{2}+\sqrt{1})+[(\sqrt{3}-\sqrt{2})(\sqrt{3}+\sqrt{2})]/(\sqrt{3}+\sqrt{2})+[(\sqrt{4}-\sqrt{3})(\sqrt{4}+\sqrt{3})]/(\sqrt{4}+\sqrt{3})+[(\sqrt{5}-\sqrt{4})(\sqrt{5}+\sqrt{4})]/(\sqrt{5}+\sqrt{4})+[(\sqrt{6}-\sqrt{5})(\sqrt{6}+\sqrt{5})]/(\sqrt{6}+\sqrt{5})+...+[(\sqrt{80}-\sqrt{79})(\sqrt{80}+\sqrt{79})]/(\sqrt{80}+\sqrt{79})+[(\sqrt{81}-\sqrt{80})(\sqrt{81}+\sqrt{80})]/(\sqrt{81}+\sqrt{80})`
`=\sqrt{2}-\sqrt{1}+\sqrt{3}-\sqrt{2}+\sqrt{4}-\sqrt{3}+\sqrt{5}-\sqrt{4}+\sqrt{6}-\sqrt{5}+...+\sqrt{80}-\sqrt{79}+\sqrt{81}-\sqrt{80}`
`=\sqrt{81}-\sqrt{1}`
`=9-1`
`=8`
`⇒2·(1/(\sqrt{1}+\sqrt{2})+1/(\sqrt{3}+\sqrt{4})+1/(\sqrt{5}+\sqrt{6})+...+1/(\sqrt{79}+\sqrt{80}))>8`
`⇒1/(\sqrt{1}+\sqrt{2})+1/(\sqrt{3}+\sqrt{4})+1/(\sqrt{5}+\sqrt{6})+...+1/(\sqrt{79}+\sqrt{80})>4`
Hãy giúp mọi người biết câu trả lời này thế nào?

Giải thích các bước giải:
Đặt `T = 1/(\sqrt{1} + \sqrt{3}) + 1/(\sqrt{5} + \sqrt{7}) + .... 1/(\sqrt{97} + \sqrt{99})`
`=> 2T = 2/(\sqrt{1} + \sqrt{3}) + 2/(\sqrt{5} + \sqrt{7}) + .... 2/(\sqrt{97} + \sqrt{99})`
Do `\sqrt{1} + \sqrt{3} < \sqrt{3} + \sqrt{5}`
`=> 1/(\sqrt{1} + \sqrt{3}) > 1/(\sqrt{3} + \sqrt{5})`
Tương tự ta có: `1/(\sqrt{5} + \sqrt{7}) > 1/(\sqrt{7} + \sqrt{9})`
.......
`1/(\sqrt{97} + \sqrt{99}) > 1/(\sqrt{99} + \sqrt{101})`
`=> 2T > 1/(\sqrt{1} + \sqrt{3}) + 1/(\sqrt{3} + \sqrt{5}) + 1/(\sqrt{5} + \sqrt{7}) + 1/(\sqrt{7} + \sqrt{9}) + .... 1/(\sqrt{97} + \sqrt{99}) + 1/(\sqrt{99} + \sqrt{101})`
`=> 2T . 2 > 2/(\sqrt{1} + \sqrt{3}) + 2/(\sqrt{3} + \sqrt{5}) + 2/(\sqrt{5} + \sqrt{7}) + 2/(\sqrt{7} + \sqrt{9}) + .... 2/(\sqrt{97} + \sqrt{99}) + 2/(\sqrt{99} + \sqrt{101})`
`=> 4T > (2(\sqrt{3} - \sqrt{1}))/((\sqrt{3} - \sqrt{1})(\sqrt{3} + \sqrt{1})) + ...... + (2(\sqrt{101} - \sqrt{99}))/((\sqrt{101} + \sqrt{99})(\sqrt{101} - \sqrt{99}))`
`=> 4T > \sqrt{3} - 1 + \sqrt{5} - \sqrt{3} + .... + \sqrt{101} - \sqrt{99}`
`=> 4T > \sqrt{101} - 1 > \sqrt{100} - 1 = 9`
`=> T > 9/4 (ĐPCM)`
Hãy giúp mọi người biết câu trả lời này thế nào?

Bảng tin
3003
32414
2695
adi đỉnh z câu này mà làm đc
4869
95322
4566
t lấy ý mạng á :))
3003
32414
2695
-)
4869
95322
4566
Kiểu có nhiều bài mò tới mốt,chỉ có bbết cái dạng mới làm dc thôi :)