chứng minh rằng B= \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+.........+\frac{1}{19}\)
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
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+....+\frac{1}{19}\)
Chứng minh rằng : B>1
Các bạn có thể giúp mk giải thật chi tiết đc ko ?
Nhiều cách lắm,ví dụ nhé:
B = ( \(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{11}\) ) + ( \(\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}\))
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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 > \(\left(\frac{1}{12}+\frac{1}{12}+...+\frac{1}{12}\right)+\left(\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)\)
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8 số 8 số
(=) B > \(\frac{8}{12}+\frac{8}{20}\)= \(\frac{2}{3}+\frac{2}{5}\)= \(\frac{16}{15}\)> 1
(=) B > 1 (đpcm)
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+..+\frac{1}{19}\),Chứng tỏ rằng : B > 1
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{4}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}=\frac{1}{4}+\frac{15}{20}=1\)
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}>\frac{1}{20}+\frac{1}{20}+....+\frac{1}{20}+\frac{1}{4}=\frac{3}{4}+\frac{1}{4}=1\)
Vậy B>1
Hok tốt
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
Chứng tỏ rằng B nhỏ hơn 1
B=1/4+(1/5+1/6+...+1/19)>1/4+15x1/20
B>1/4+15/20=1/4+3/4=1
\(\Rightarrow\)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 \(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{19}>1\)
\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+....+\frac{1}{19}\)Chứng tỏ rằng B > 1
B = \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
B = \(\left(\frac{1}{4}+\frac{1}{5}+...+\frac{1}{11}\right)+\left(\frac{1}{12}+\frac{1}{13}+...+\frac{1}{19}\right)>\left(\frac{1}{11}+...+\frac{1}{11}\right)+\left(\frac{1}{19}+...+\frac{1}{19}\right)\)
B > \(\frac{240}{209}\)
Vậy B > 1.
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\)
CHỨNG MINH RẰNG B > 1
các bạn giải đầy đủ hộ mình nhớ giải thích rõ ràng nhé mình cảm ơn
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
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
B=1/4+1/5+1/6+...+1/19
B=1/4+(1/5+1/6+1/7+...+1/9)+(1/10+1/11+...+1/19)>1/4+(1/9+1/9+...+1/9)+(1/19+1/19+...+1/19)
B>1/4+5/9+10/19>1/4+1/2+1/2>1(vì 5/9>4/8=1/2;10/19>9/18=1/2)
Cho B= \(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+...+\frac{1}{19}\). Chứng minh B>1
Ta có :
\(B=\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+..............+\frac{1}{19}\)
\(B=\frac{1}{4}+\left(\frac{1}{5}+\frac{1}{6}+.....+\frac{1}{9}\right)+\left(\frac{1}{10}+\frac{1}{11}+.........+\frac{1}{19}\right)\)
Ta thấy :
\(\frac{1}{5}+\frac{1}{6}+...+\frac{1}{9}>\frac{1}{9}+\frac{1}{9}+...+\frac{1}{9}=\frac{1}{9}.5=\frac{5}{9}>\frac{1}{2}\)
\(\frac{1}{10}+\frac{1}{11}+....+\frac{1}{19}>\frac{1}{19}+\frac{1}{19}+...+\frac{1}{19}=\frac{1}{19}.5>\frac{10}{19}>\frac{1}{2}\)
\(\Rightarrow B>\frac{1}{4}+\frac{1}{2}+\frac{1}{2}>1\)
ta có
1/4+1/5+....+1/10>1/10,7=7/10
1/11+1/12+.....+1/19>1/20,9=9/20
kết hợp lại ta có b=1/4+1/5+1/6+......+1/19>7/10+9/20=23/20>1
vậy b>1
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