Chứng tỏ rằng
a, 11/15 < 1/21 + 1/22 + 1/23 +................+ 1/59 + 1/6 < 3/2
b, 3 < 1 + 1/2 + 1/3+ 1/4 +..................+ 1/63<6
Bài 1 :Chứng tỏ rằng :
a) \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)
b) \(3< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
Câu hỏi của Hoàng Đỗ Việt - Toán lớp 6 | Học trực tuyến
Bài 1 :
Ta có;\(\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{30}>\frac{1}{30}.10=\frac{1}{3}\)
\(\frac{1}{31}+\frac{1}{32}+...+\frac{1}{60}>\frac{1}{60}.30>\frac{1}{30}.24=\frac{2}{5}\)
Do đó :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}>\frac{1}{3}+\frac{2}{5}=\frac{11}{15}\left(1\right)\)
Mặt khác :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}.20=1\)
\(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}.20=\frac{1}{2}\)
Do đó :
\(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{60}< 1+\frac{1}{2}=\frac{3}{2}\left(2\right)\)
Từ (1 ) và (2) ta suy ra điều phải chứng minh
Bài 2 :
Đặt \(S=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}\)
MỘT MẶT ,TA CÓ THỂ VIẾT
\(S=\left(1+\frac{1}{2}\right)+\left(\frac{1}{3}+\frac{1}{4}\right)+\left(\frac{1}{5}+\frac{1}{6}+\frac{1}{7}+\frac{1}{8}\right)\)\(+\left(\frac{1}{9}+\frac{1}{10}+...+\frac{1}{16}\right)+\left(\frac{1}{17}+\frac{1}{18}+...+\frac{1}{32}\right)\)\(+\left(\frac{1}{33}+\frac{1}{34}+...+\frac{1}{63}+\frac{1}{64}\right)-\frac{1}{64}\)
\(>\frac{1}{2}.2+\frac{1}{4}.2+\frac{1}{8}.4+\frac{1}{16}.8+\frac{1}{32}.16+\frac{1}{64}.32-\frac{1}{64}\)\(=\frac{7}{2}-\frac{1}{64}=\frac{223}{64}>\frac{192}{64}=3\left(1\right)\)
Mặt khác ,ta lại có\(S=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)\)\(+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)\)\(+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)< \)\(1+\frac{1}{2}.2+\frac{1}{4}.4+\frac{1}{8}.8+\frac{1}{16}.16+\frac{1}{32}.32=6\left(2\right)\)
Từ (1) và (2 ) ta kết luận \(3< S< 6\)
Chúc bạn học tốt ( -_- )
a) Đặt \(A=\frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}\)
Ta có:
\(A=\left(\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}\right)+\left(\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}\right)\)
+ Vì \(\frac{1}{21}>\frac{1}{40};\frac{1}{22}>\frac{1}{40};...;\frac{1}{40}=\frac{1}{40}\)
\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}>\frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)( 20 phân số \(\frac{1}{40}\)) \(=20.\frac{1}{40}=\frac{1}{2}.\)
+ Vì \(\frac{1}{41}>\frac{1}{60};\frac{1}{42}>\frac{1}{60};...;\frac{1}{60}=\frac{1}{60}\)
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}>\frac{1}{60}+\frac{1}{60}+...+\frac{1}{60}\)( 20 phân số \(\frac{1}{60}\)) \(=20.\frac{1}{60}=\frac{1}{3}\)
\(\Rightarrow A>\frac{1}{2}+\frac{1}{3}=\frac{5}{6}=\frac{75}{90}>\frac{66}{90}=\frac{11}{15}\)
\(\Rightarrow A>\frac{11}{15}\left(1\right)\)
Lại có:
+ Vì \(\frac{1}{21}< \frac{1}{20};\frac{1}{22}< \frac{1}{20};...;\frac{1}{40}< \frac{1}{20}\)
\(\Rightarrow\frac{1}{21}+\frac{1}{22}+...+\frac{1}{40}< \frac{1}{20}+\frac{1}{20}+...+\frac{1}{20}\)( 20 phân số \(\frac{1}{20}\)) \(=20.\frac{1}{20}=1\)
+ Vì \(\frac{1}{41}< \frac{1}{40};\frac{1}{42}< \frac{1}{40};...;\frac{1}{60}< \frac{1}{40}\)
\(\Rightarrow\frac{1}{41}+\frac{1}{42}+...+\frac{1}{60}< \frac{1}{40}+\frac{1}{40}+...+\frac{1}{40}\)( 20 phân số \(\frac{1}{40}\)) \(=20.\frac{1}{40}=\frac{1}{2}\)
\(\Rightarrow A< 1+\frac{1}{2}=\frac{3}{2}\)
\(\Rightarrow A< \frac{3}{2}\left(2\right)\)
Từ\(\left(1\right)\);\(\left(2\right)\) \(\Rightarrow\frac{11}{15}< A< \frac{3}{2}\left(đpcm\right).\)
b) Đặt \(B=1+\frac{1}{2}+\frac{1}{3}+...+\frac{1}{63}\)
Ta có:
\(B=1+\left(\frac{1}{2}+\frac{1}{3}\right)+\left(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}\right)+\left(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}\right)\)\(+\left(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}\right)+\left(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}\right)\)
+ \(1=1\)
+\(\frac{1}{2}+\frac{1}{3}< \frac{1}{2}+\frac{1}{2}=1\)
+\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}< \frac{1}{4}+\frac{1}{4}+\frac{1}{4}+\frac{1}{4}=1\)
+\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}< \frac{1}{8}+\frac{1}{8}+...+\frac{1}{8}=1\)
Tương tự ta được:
+\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}< 1\)
+\(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}< 1\)
\(\Rightarrow A< 1+1+1+1+1+1=6\left(1\right)\)
Lại có:
+\(1=1\)
+\(\frac{1}{2}+\frac{1}{3}>\frac{1}{3}+\frac{1}{3}=\frac{2}{3}\)
+\(\frac{1}{4}+\frac{1}{5}+\frac{1}{6}+\frac{1}{7}>\frac{1}{7}+\frac{1}{7}+\frac{1}{7}+\frac{1}{7}=\frac{4}{7}\)
+\(\frac{1}{8}+\frac{1}{9}+...+\frac{1}{15}>\frac{1}{15}+\frac{1}{15}+...+\frac{1}{15}=\frac{8}{15}\)
Tương tự, ta được:
+\(\frac{1}{16}+\frac{1}{17}+...+\frac{1}{31}>\frac{16}{31}\)
+\(\frac{1}{32}+\frac{1}{33}+...+\frac{1}{63}< \frac{32}{63}\)
\(\Rightarrow A>1+\frac{2}{3}+\frac{4}{7}+\frac{8}{15}+\frac{16}{31}+\frac{32}{63}\)\(=1+\frac{18}{15}+\frac{64}{63}+\frac{16}{31}>1+\frac{15}{15}+\frac{63}{63}=3\left(2\right)\)
Từ\(\left(1\right)\)và \(\left(2\right)\Rightarrow3< A< 6\left(đpcm\right).\)
Chứng tỏ rằng: 11/15 < 1/21 + 1/22 + 1/23 + ... + 1/59 + 1/60 < 3/2
Chứng tỏ rằng: 11/15 < 1/21 + 1/22 + 1/23 + ... + 1/59 + 1/60 < 3/2
Cho S = 1/21 + 1/22 + 1/23 +... + 1/60
S1=1/21 + 1/22 +..+ 1/40 (20 số hạng); S2= 1/41 + 1/42 +... + 1/60 (20 số hạng)
* Ta thấy: S1 > 1/40 x 20 = 1/2 (vì 1/40 = 1/40, 19 số hạng kia đều lớn hơn 1/40); S2 > 1/60 x 20 = 1/3
\(\Rightarrow\)S > 1/2 + 1/3 = 5/6 = 25/30 > 22/30 = 11/15
Vậy 1/21 + 1/22 + ... + 1/60 > 11/15
* Ta thấy: S1 < 1/21 x 20 = 20/21(vì 1/20 = 1/20, 19 số hạng còn lại đều bé hơn 1/21); S2 < 1/41 x 20 = 20/41
\(\Rightarrow\)S < 20/21 + 20/41 = 1240/861 < 3/2 (đoạn này thì bạn phải dùng máy tính chứ mik ko bt tính nhanh kiểu j)
Ta có đpcm
Hãy chứng tỏ rằng:
a) 1/41+1/42+1/43+...+1/79+1/80>7/12
b)11/15<1/21+1/22+1/23+...+1/59+1/60<3/2
a) Chứng minh: \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{60}< \frac{3}{2}\)
b) Chứng minh: \(3< 1+\frac{1}{2}+\frac{1}{3}+\frac{1}{4}+...+\frac{1}{63}< 6\)
Y Ribi Nkok Ngok Lê Nguyễn Ngọc Nhi Lê Anh Duy Nguyễn Thị Diễm Quỳnh trần thị diệu linh kudo shinichi Nguyen Giang Thủy Tiên Nguyễn Việt Lâm
Chứng minh
11/15<1/21+1/22+1/23+....+1/59+1/60<3/2
Chứng minh rằng:
11/15 < 1/21 + 1/22 + 1/23+.....+1/58 + 1/59 + 1/60 < 3/2
Đặt A=1/21+1/22+...+1/60=(1/21+1/22+...+1/40)+(1/41+1/42+...+1/60)
Ta có:1/21>1/40, 1/22>1/40,..., 1/39>1/40
=>1/21+1/226+...+1/40>1/40+1/40+...+1/40=1/40.20=1/2
1/41>1/60, 1/42>1/60,...,1/59>1/60
=>1/41+1/42+...+1/60>1/60+1/60+...+1/60=1/60.20=1/3
=>1/21+1/22+...+1/60>1/2+1/3=5/6>11/15
=>A>11/15 (1)
Lại có: 1/21<1/20, 1/22<1/20,...,1/40<1/20
=>1/21+1/22+...+1/40<1/20+1/20+...+1/20=1/20.20=1
1/41<1/40, 1/42<1/40,...,1/60<1/40
=>1/41+1/42+...+1/60<1/40+1/40+...+1/40=1/40.20=1/2
=>1/21+1/22+...+1/60<1+1/2=3/2
=>A<3/2 (2)
Từ (1) và (2)
=>11/15<A<3/2
=>11/15<1/21+1/22+...+1/60<3/2 (đpcm)
Chứng minh
a, 1/4+1/16+1/36+1/64+1/100+1/144+1/196<1/2
b, 11/15<1/21+1/22+1/23+...+1/59+1/60<3/2
a,\( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+ \dfrac{1}{64}+ \dfrac{1}{100}+ \dfrac{1}{144}+ \dfrac{1}{196}\)
= \( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+...+ \dfrac{1}{196} < \dfrac{1}{2^2-1}+ \dfrac{1}{4^2-1}+ \dfrac{1}{6^2-1}+...+ \dfrac{1}{14^2-1}\)
= \( \dfrac{1}{1.3}+ \dfrac{1}{3.5}+ \dfrac{1}{5.7}+...+ \dfrac{1}{13.15}\)
= \( \dfrac{1}{2}(1- \dfrac{1}{3}+ \dfrac{1}{3}- \dfrac{1}{5}+ \dfrac{1}{5}- \dfrac{1}{7}+ \dfrac{1}{7}-...- \dfrac{1}{13}+ \dfrac{1}{13}- \dfrac{1}{15})\)
= \( \dfrac{1}{2}(1- \dfrac{1}{15})< \dfrac{1}{2}\)
Vậy \( \dfrac{1}{4}+ \dfrac{1}{16}+ \dfrac{1}{36}+ \dfrac{1}{64}+ \dfrac{1}{100}+ \dfrac{1}{144}+ \dfrac{1}{196}\) \(<\dfrac{1}{2} \)
b,A= \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)
\(=(\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+....+\dfrac{1}{40})+(\dfrac{1}{41}+...+1...\)
\(=(\dfrac{20}{20.21}+\dfrac{21}{21.22}+...+\dfrac{39}{39.40})+(40/...\)
\(20(\dfrac{1}{20.21}+\dfrac{1}{21.22}+...\dfrac{1}{39.40})+40(\dfrac{1}{40}...\)
\(20(\dfrac{1}{20}-\dfrac{1}{40})+40(\dfrac{1}{40}-\dfrac{1}{60})>\dfrac{11}{15}\)
Lại có \(A<40(\dfrac{1}{20.21}+...\dfrac{1}{39.40})+60(\dfrac{1}{40.41}+...+...\)
\(=40(\dfrac{1}{20}-\dfrac{1}{40})+60(\dfrac{1}{40}-\dfrac{1}{60})<\dfrac{3}{2}\)
=> \(\dfrac{11}{15}<\dfrac{1}{21}+\dfrac{1}{22}+\dfrac{1}{23}+...+\dfrac{1}{59}+\dfrac{1}{60}<\dfrac{3}{2}\)
Chứng minh rằng: \(\frac{11}{15}< \frac{1}{21}+\frac{1}{22}+\frac{1}{23}+...+\frac{1}{59}+\frac{1}{60}< \frac{3}{2}\)