cho A = \(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{58}+\frac{1}{59}\)chứng minh A <\(\frac{3}{2}\)
Những câu hỏi liên quan
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 ^-^
Chứng minh rằng: \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)
Chứng minh
\(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)
Đặt \(C=\frac{1}{21}+\frac{1}{22}+....+\frac{1}{60}=\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)\)
Ta có: \(\frac{1}{21}>\frac{1}{40};\frac{1}{22}>\frac{1}{40};....\frac{1}{39}>\frac{1}{40}\)
\(\Rightarrow\frac{1}{21}+\frac{1}{22}+....+\frac{1}{39}+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{40}.20=\frac{1}{2}\)
\(\frac{1}{41}>\frac{1}{60};\frac{1}{42}>\frac{1}{60};...\frac{1}{59}>\frac{1}{60}\)
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}=\frac{1}{60}.20=\frac{1}{3}\)
\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}>\frac{11}{15}\)
Vậy \(C>\frac{11}{15}\) (1)
Lại có: \(\frac{1}{21}< \frac{1}{20};\frac{1}{22}< \frac{1}{20};...\frac{1}{40}< \frac{1}{20}\)
\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}+....+\frac{1}{20}=\frac{1}{20}.20=1\)
\(\frac{1}{41}< \frac{1}{40};\frac{1}{42}< \frac{1}{40};...\frac{1}{60}< \frac{1}{40}\)
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}=\frac{1}{40}.20=\frac{1}{2}\)
\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< \frac{1}{2}+1=\frac{3}{2}\)
Vậy \(C< \frac{3}{2}\) (2)
Từ (1) và (2) suy ra \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< \frac{3}{2}\)
Đúng 1
Bình luận (0)
Chứng minh
\(\frac{11}{15} < \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+....+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)
1)
\(Cho:\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{200}\)
Chứng minh: \(A>\frac{9}{10}\)
2)
Cho \(B=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
Chứng minh \(B>\frac{7}{12}\)
1)
Cho \(\frac{1}{20}+\frac{1}{21}+\frac{1}{22}+...+\frac{1}{200}\)
Chứng minh: \(A>\frac{9}{10}\)
2)
Cho \(B=\frac{1}{101}+\frac{1}{102}+\frac{1}{103}+...+\frac{1}{200}\)
Chứng minh: \(B>\frac{7}{12}\)
choA=\(\frac{1}{21}+\frac{1}{22}+.........+\frac{1}{59}+\frac{1}{60}\)
chứng tỏ \(\frac{11}{15}
Chứng minh:
c.\(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)
b.\(\frac{1}{5}+\frac{1}{13}+\frac{1}{25}+\frac{1}{41}+\frac{1}{61}+\frac{1}{85}+\frac{1}{113}< \frac{1}{2}\)
a.\(\frac{1}{4}+\frac{1}{16}+\frac{1}{36}+\frac{1}{64}+\frac{1}{100}+\frac{1}{144}+\frac{1}{196}< \frac{1}{2}\)
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
Đúng 0
Bình luận (0)
GIÚP NHA , AI LÀM ĐƯƠC 1 NGÀY TK 3TK
Đúng 0
Bình luận (0)
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>\(\frac{9}{10}\)
Đúng 0
Bình luận (0)