CHO 1/5+1/13+1/14+1/15+1/61+1/62+1/63
cHỨNG minh 3/7<S<1/2
s= 1/5+1/13+1/14+1/15+1/61+1/62+1/63.CMR:3/7<S<1/2
chứng minh rằng s=1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/2
Chứng minh S = 1/5 +1/13+ 1/14+ 1/15 +1/61+ 1/62 <1/2
Ta có:
\(\frac{1}{5}=\frac{1}{5}\)
\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}
chứng minh 1/5+1/13+1/14+1/15+1/61+1/62+1/63 < 1/2
Ta có :
S = \(\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)
Chứng minh S = 1/5 +1/13+ /14+1/15+1/61+1/62+1/63 < 1/2
Ta có:
\(\frac{1}{5}=\frac{1}{5}\)
\(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}
Ta có: \(S=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\right)
CHỨNG MINH 1-(1/5 +1/13 +1/14+1/15+1/61 +1/62 +1/63) > 1/2
\(đpcm\Leftrightarrow\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}
chứng minh rằng: S=1/5+1/13+1`/14+1/15+1/61+1/62+1/63<1/2
\(\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}
Ta có:
S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)<1/5+1/12.3+1/60.3
=>S<1/5+1/4+1/20=10/20
Hay S<1/2
chung minh rang :
S= 1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/2
S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)
(*)Ta có:
1/13<1/12
1/14<1/12
1/15<1/12
=>1/13+1/14+1/15<1/12
(*)Ta lại có:
1/61<1/60
1/62<1/60
1/63<1/60
=>1/61+1/62+1/63<1/60
=>S<1/5+1/12.3+1/60.3
S<1/5+1/4+1/20
S<1/2
S=1/5+(1/13+1/14+1/15)+(1/61+1/62+1/63)
(*)Ta có:
1/13<1/12
1/14<1/12
1/15<1/12
=>1/13+1/14+1/15<1/12
(*)Ta lại có:
1/61<1/60
1/62<1/60
1/63<1/60
=>1/61+1/62+1/63<1/60
=>S<1/5+1/12.3+1/60.3
S<1/5+1/4+1/20
S<1/2
mọi người để ý cau trả lời của bạn Dung và đứa chép bài Trà My nhé vì
1/13<1/12
1/14<1/12
1/15<1/12
nên 1/13+1/14+1/15<1/12(đây là câu trả lời của Dung và đứa chép bài)
bạn sai vì 1/13<1/12
1/14<1/12
1/15<1/12
nhưng chưa chắc 1/13+1/14+1/15<1/12(mình chưa tính nhé ^^) và từ đó ta cũng kết luận đc Trà My là 1 đứa chép bài
Chứng minh
a=1/5+1/13+1/14+1/15+1/61+1/62+1/63<1/2
Ta có: \(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}\)
\(A=\frac{1}{5}+\left(\frac{1}{13}+\frac{1}{14}+\frac{1}{15}\right)+\left(\frac{1}{62}+\frac{1}{62}+\frac{1}{63}\right)\)
\(A=\frac{1}{5}+\frac{1}{15}.3+\frac{1}{63}.3\)
\(A=\frac{1}{5}+\frac{1}{5}+\frac{1}{21}\)
\(A=\frac{47}{105}\)
Mà: \(\frac{47}{105}< \frac{47}{94}=\frac{1}{2}\)
Nên \(A=\frac{1}{5}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{61}+\frac{1}{62}+\frac{1}{63}< \frac{1}{2}\)