Cho B= \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\). Chứng minh B>1
Cho \(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\) .Chứng minh rằng B>1
B = 1/4 + 1/5 + 1/6 + ... + 1/19 > 1
B = 1/4+﴾1/5+1/6+...+1/9﴿+﴾1/10+1/11+...+1/19﴿
Vì 1/5+1/6+...+1/9 > 1/9+1/9+...+1/9 nên 1/5+1/6+...+1/9 > 5/9 >1/2
Vì 1/10+1/11+...+1/19 > 1/19+1/19+...+1/19 nên 1/10+1/11+...+1/19 > 10/19 >1/2
Suy ra: B > 1/4+1/2+1/2 > 1
chứng minh rằng B= \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+.........+\frac{1}{19}\)
B>1
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{16}+\frac{1}{16}+\frac{1}{16}+...+\frac{1}{16}=\frac{16}{16}=1\)
Cho B=\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+........+\frac{1}{19}\) . hãy chứng minh B>1
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}=\frac{1}{4}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\left(\frac{1}{5}+...+\frac{1}{8}\right)+\left(\frac{1}{9}+...+\frac{1}{16}\right)\)
\(\frac{1}{5}+...+\frac{1}{8}>\frac{1}{8}.4=\frac{1}{2}\)
\(\frac{1}{9}+...+\frac{1}{16}\frac{1}{2}+\frac{1}{2}=1\)
\(SuyraB>1\)
Ta có: \(B=\left(\frac{1}{4}+\frac{1}{19}\right).8\)
\(B=2\frac{8}{19}\)
=> B>1
cho B = \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{19}.\)
- Hãy chứng tỏ rằng B > 1
\(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{19}=\frac{1}{4}+\left(\frac{1}{5}+...+\frac{1}{9}\right)+\left(\frac{1}{10}+...+\frac{1}{19}\right)\) > \(\frac{1}{4}+\left(\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}\right)+\left(\frac{1}{19}+...+\frac{1}{19}\right)\)> \(\frac{1}{4}+\frac{5}{9}+\frac{10}{19}>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}=1\)
Vậy \(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{19}>1\)
Bài 5: Cho B = \(\frac{1}{4} +\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\). Hãy chứng tỏ B > 1.
cho B =\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\),hãy chứng minh B>1
1)Cho A=\(\frac{196}{197}\)+\(\frac{197}{198}\)
B=\(\frac{196+197}{197+198}\)
So sánh A và B
2)Cho B=\(\frac{1}{4}\)+\(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+...+\frac{1}{19}\).Chứng minh B>1
3)Tính nhanh
\(\frac{\frac{3}{5}+\frac{3}{7}-\frac{3}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{11}}\)
3)
3/5 + 3/7-3/11 / 4/5 + 4/7- 4/11
= 3.( 1/5 + 1/7 - 1/11)/4.(1/5+1/7-1/11)
= 3/4
1,
ta có B = 196+197/197+198 = 196/(197+198) + 197/(197+198)
196/197 > 196/197+198
197/198 > 197/197+198
=> A>B
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+....+\frac{1}{19}\)
Chứng minh rằng : B>1
Các bạn có thể giúp mk giải thật chi tiết đc ko ?
Nhiều cách lắm,ví dụ nhé:
B = ( \(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{11}\) ) + ( \(\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}\))
______________________ _________________________
B C
-Ta xét B ( vì bạn bảo chi tiết nên tôi làm như vậy còn ở bài thì không cần như vậy )
\(\frac{1}{4}>\frac{1}{12}\);...; \(\frac{1}{11}>\frac{1}{12}\)
-Xét C : \(\frac{1}{12}>\frac{1}{20};...;\frac{1}{19}>\frac{1}{20}\)
(=) B > \(\left(\frac{1}{12}+\frac{1}{12}+...+\frac{1}{12}\right)+\left(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)\)
_________________ ___________________
8 số 8 số
(=) B > \(\frac{8}{12}+\frac{8}{20}\)= \(\frac{2}{3}+\frac{2}{5}\)= \(\frac{16}{15}\)> 1
(=) B > 1 (đpcm)
Cho B=\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
Hãy chứng tỏ rằng B>1
\(B=\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{11}\right)+\left(\frac{1}{12}+...+\frac{1}{19}\right)>\left(\frac{1}{12}+\frac{1}{12}+...+\frac{1}{12}\right)+\left(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)\)=> \(B>\frac{8}{12}+\frac{8}{20}=\frac{2}{3}+\frac{2}{5}=\frac{16}{15}>\frac{15}{15}=1\)
=> ĐPCM
mình có bài làm giống cô Trần Thị Loan
tk mình nhé