\(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{49}<\frac{13}{12}\)
C/m ::
\(S=\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}+\frac{1}{21}+\frac{1}{22}>\frac{1}{2}\)
Cho A=\(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{58}+\frac{1}{59}.Chứngtỏ\)rằng A<\(\frac{3}{2}\)
ta có
A=1/20 + 1/21+1/22+....+1/59
=(1/20+1/21+...+1/39)+(1/40+1//41+....+1/59)<1/20.20+1/40.20=1 + 1/2=3/2
vậy A<3/2
Chúc bạn học tốt nha ^-^
Chưng Minh:\(\frac{1}{21}\)+\(\frac{1}{22}\)+\(\frac{1}{23}\)+....+\(\frac{1}{49}\)+\(\frac{1}{50}\)<\(1\frac{1}{12}\)
(1/21+1/22+...+1/30)+(1/31+...+1/40)+(1/41+...+1/50)
(1/21+1/22+...+1/30)<1/20+..+1/20=1/20*10=1/2
(1/31+...+1/40)<1/30+..+1/30=1/30*10=1/3
(1/41+...+1/50)<1/40+...+1/40=1/40*10=1/4
Suy ra day so <1/2+1/3+1/4=13/12=1/1/12=>dpcm
k cho minh nhe
cho A = \(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{58}+\frac{1}{59}\)chứng minh A <\(\frac{3}{2}\)
Ta có \(A=\left(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{39}\right)+\left(\frac{1}{40}+\frac{1}{41}+...+\frac{1}{59}\right)\)
\(A< \left(\frac{1}{20}+\frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\right)+\left(\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\right)\)
\(A< \frac{20}{20}+\frac{20}{40}\)
\(A< \frac{3}{2}\)
CHO
S=\(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{199}+\frac{1}{200}\)
CHỨNG MINH RẰNG S>\(\frac{9}{10}\)
S = \(\frac{1}{20}+\frac{1}{21}...+\frac{1}{199}+\frac{1}{200}\) ( có 181 phân số )
=> S > \(\frac{1}{200}+\frac{1}{200}+...+\frac{1}{200}+\frac{1}{200}\)
=> S > \(\frac{1}{200}.181\)
=> S > \(\frac{181}{200}\)> \(\frac{180}{200}\)= \(\frac{9}{10}\)
Vậy S > 9 / 10
GIÚP NHA , AI LÀM ĐƯƠC 1 NGÀY TK 3TK
S = \(\frac{1}{20}\)+ \(\frac{1}{21}\)+ ....+\(\frac{1}{200}\)có 181 p/s
mà \(\frac{1}{20}\)>\(\frac{1}{200}\)
.............
\(\frac{1}{199}\)>\(\frac{1}{200}\)
\(\frac{1}{200}\)=\(\frac{1}{200}\)
nên ta có S > \(\frac{1}{200}\)+ \(\frac{1}{200}\)+..... có 181 phân số \(\frac{1}{200}\)
vậy \(\frac{1}{200}\)*181=\(\frac{181}{200}\)mà \(\frac{181}{200}\)>\(\frac{9}{10}\)mà \(\frac{1}{20}\)+......+\(\frac{1}{200}\)(có 181 số)>\(\frac{1}{200}\)+\(\frac{1}{200}\)(có 181 p/s \(\frac{1}{200}\))>\(\frac{9}{10}\)
Vậy ==> S>\(\frac{9}{10}\)
Chứng minh rằng:\(\frac{1}{20!}+\frac{1}{21!}+\frac{1}{22!}+..........+\frac{1}{5000!}\)<\(\frac{1}{19!}\)
Thực hiện so sánh:\(\frac{1}{12}+\frac{1}{13}+\frac{1}{14}+\frac{1}{15}+\frac{1}{16}+\frac{1}{17}\)\(+\frac{1}{18}+\frac{1}{19}+\frac{1}{20}+\frac{1}{21}+\frac{1}{22}\)\(+\frac{1}{23}\)với \(\frac{5}{6}\)
Đặt S=1/12+1/13+1/14+1/15+...+1/23
ta có 1/12+1/13+1/14+1/15+...+1/22+1/23 = (1/12+1/13+1/14+...+1/17)+(1/18+1/19+...+1/23)
đặt A=1/12+1/13+1/14+...+1/17
ta có
1/13<1/12
1/14<1/12
..........................
.........................
1/17<1/12
=>A<1/12+1/12+1/12+....+1/12 (có 6 phân số)
=>A<1x6/12
=>A<1/2 (1)
Đặt B=1/18+1/19+...+11/23
ta có
1/19<1/18
1/20<1/18
...........................
..........................
1/23<1/18
=> B<1/18+1/18+1/18+...+1/18 (có 6 phân số)
=>B<1x 6/18
=>B<1/3 (2)
từ 1 và 2 =>S=A+B<1/2+1/3
=>S<5/6 (dpcm)
k cho mình nhé
Đặt S=1/12+1/13+1/14+1/15+...+1/23
ta có 1/12+1/13+1/14+1/15+...+1/22+1/23 = (1/12+1/13+1/14+...+1/17)+(1/18+1/19+...+1/23)
đặt A=1/12+1/13+1/14+...+1/17
ta có
1/13<1/12
1/14<1/12
..........................
.........................
1/17<1/12
=>A<1/12+1/12+1/12+....+1/12 (có 6 phân số)
=>A<1x6/12
=>A<1/2 (1)
Đặt B=1/18+1/19+...+11/23
ta có
1/19<1/18
1/20<1/18
...........................
..........................
1/23<1/18
=> B<1/18+1/18+1/18+...+1/18 (có 6 phân số)
=>B<1x 6/18
=>B<1/3 (2)
từ 1 và 2 =>S=A+B<1/2+1/3
=>S<5/6 (dpcm)
k cho mình nhé
Chứng minh rằng :
\(\frac{7}{12}< \frac{1}{21}+\frac{1}{20}+...+\frac{1}{40}< 1\)
Chú ý p/s thứ 2 là 1/20 chứ k phải 1/22 nha
Cho \(S=\frac{5}{20}+\frac{5}{21}+\frac{5}{22}+...+\frac{5}{49}\).CMR: 3<S<8