1, Chứng minh
a) A=\(\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
b) B=\(\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+\dfrac{36}{5.7.9}+....+\dfrac{36}{25.26.27}< 3\)
Những câu hỏi liên quan
Chứng minh:
Adfrac{1}{1.2.3}+dfrac{1}{2.3.4}+dfrac{1}{3.4.5}+...+dfrac{1}{18.19.20} dfrac{1}{4}
Bdfrac{36}{1.3.5}+dfrac{36}{5.7.9}+dfrac{36}{9.11.13}+...+dfrac{36}{25.27.29} 3
Cdfrac{1}{2^2}+dfrac{1}{3^2}+dfrac{1}{4^2}+...+dfrac{1}{n^2}in 1left(nin N,nge2right)
Ddfrac{1}{4^2}+dfrac{1}{6^2}+dfrac{1}{8^2}+...+dfrac{1}{left(2nright)^2} 4left(nin N,nge2right)
Edfrac{2!}{3!}+dfrac{2!}{4!}+dfrac{2!}{5!}+...+dfrac{2!}{n!} 1left(nin N,nge3right)
Đọc tiếp
Chứng minh:
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
\(B=\dfrac{36}{1.3.5}+\dfrac{36}{5.7.9}+\dfrac{36}{9.11.13}+...+\dfrac{36}{25.27.29}< 3\)
\(C=\dfrac{1}{2^2}+\dfrac{1}{3^2}+\dfrac{1}{4^2}+...+\dfrac{1}{n^2}\in< 1\left(n\in N,n\ge2\right)\)
\(D=\dfrac{1}{4^2}+\dfrac{1}{6^2}+\dfrac{1}{8^2}+...+\dfrac{1}{\left(2n\right)^2}< 4\left(n\in N,n\ge2\right)\)
\(E=\dfrac{2!}{3!}+\dfrac{2!}{4!}+\dfrac{2!}{5!}+...+\dfrac{2!}{n!}< 1\left(n\in N,n\ge3\right)\)
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+....+\dfrac{1}{18.19.20}=\dfrac{1}{2}\left(\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{2}\left(\dfrac{1}{2}-\dfrac{1}{19.20}\right)\\ =\dfrac{1}{4}-\dfrac{1}{2.19.20}< \dfrac{1}{4}\)
Cái B TT nhé
\(\dfrac{1}{2^2}+\dfrac{1}{3^2}+....+\dfrac{1}{n^2}< \dfrac{1}{1.2}+\dfrac{1}{2.3}+...+\dfrac{1}{\left(n-1\right)n}\\ =1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{n-1}-\dfrac{1}{n}\\ =1-\dfrac{1}{n}< 1\)
D TT
E mk thấy nó ss ớ
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15 tháng 4 2023 lúc 13:47
a) dfrac{2}{1.2.3}+dfrac{2}{2.3.4}+dfrac{2}{3.4.5}+...+dfrac{2}{18.19.20}b) dfrac{4}{1.3.5}+dfrac{4}{3.5.7}+dfrac{4}{5.7.9}+...+dfrac{4}{21.23.25}c) dfrac{3}{1.2}-dfrac{5}{2.3}+dfrac{7}{3.4}-dfrac{9}{4.5}+...+dfrac{39}{19.20}-dfrac{41}{20.21}d) dfrac{8}{9}cdotdfrac{15}{16}cdotdfrac{24}{25}cdot...cdotdfrac{99}{100}cdotdfrac{120}{121}e) left(1+dfrac{7}{9}right)left(1+dfrac{7}{20}right)left(1+dfrac{7}{33}right)left(1+dfrac{7}{48}right)...left(1+dfrac{7}{180}right)Các bạn không nhất thiết phải làm h...
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a) \(\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{18.19.20}\)
b) \(\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+\dfrac{4}{5.7.9}+...+\dfrac{4}{21.23.25}\)
c) \(\dfrac{3}{1.2}-\dfrac{5}{2.3}+\dfrac{7}{3.4}-\dfrac{9}{4.5}+...+\dfrac{39}{19.20}-\dfrac{41}{20.21}\)
d) \(\dfrac{8}{9}\cdot\dfrac{15}{16}\cdot\dfrac{24}{25}\cdot...\cdot\dfrac{99}{100}\cdot\dfrac{120}{121}\)
e) \(\left(1+\dfrac{7}{9}\right)\left(1+\dfrac{7}{20}\right)\left(1+\dfrac{7}{33}\right)\left(1+\dfrac{7}{48}\right)...\left(1+\dfrac{7}{180}\right)\)
Các bạn không nhất thiết phải làm hết, làm cho nó dễ hiểu được thì càng tốt để mk vận dụng
a: \(=\dfrac{1}{1\cdot2}-\dfrac{1}{2\cdot3}+\dfrac{1}{2\cdot3}-\dfrac{1}{3\cdot4}+...+\dfrac{1}{18\cdot19}-\dfrac{1}{19\cdot20}\)
=1/2-1/380
=179/380
b: \(=\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{21\cdot23}-\dfrac{1}{23\cdot25}\)
\(=\dfrac{1}{3}-\dfrac{1}{575}=\dfrac{572}{1725}\)
c: \(=1+\dfrac{1}{2}-\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{19}+\dfrac{1}{20}-\dfrac{1}{20}-\dfrac{1}{21}\)
=1-1/21
=20/21
d: \(=\left(1-\dfrac{1}{9}\right)\left(1-\dfrac{1}{16}\right)\cdot...\cdot\left(1-\dfrac{1}{121}\right)\)
\(=\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot...\cdot\dfrac{10}{11}\cdot\dfrac{3}{2}\cdot\dfrac{4}{3}\cdot...\cdot\dfrac{12}{11}\)
\(=\dfrac{2}{11}\cdot\dfrac{12}{2}=\dfrac{12}{11}\)
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CMR \(B=\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+\dfrac{36}{5.7.9}+...+\dfrac{36}{25.27.29}< 3\)
Chứng minh rằng:B=\(\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+\dfrac{36}{5.7.9}+...+\dfrac{36}{25.27.29}< 3\)
Ai làm đúng mình tick cho nha
Ta có :
\(B=\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+\dfrac{36}{5.7.9}+...............+\dfrac{36}{25.27.29}\)
\(B=9\left(\dfrac{4}{1.3.5}+\dfrac{4}{3.5.7}+\dfrac{4}{5.7.9}+.............+\dfrac{4}{25.27.29}\right)\)
\(B=9\left(\dfrac{1}{1.3}-\dfrac{1}{3.5}+\dfrac{1}{3.5}-\dfrac{1}{5.7}+\dfrac{1}{5.7}-\dfrac{1}{7.9}+...........+\dfrac{1}{25.27}-\dfrac{1}{27.29}\right)\)
\(B=9\left(\dfrac{1}{1.3}-\dfrac{1}{27.29}\right)\)
\(B=9\left(\dfrac{1}{3}-\dfrac{1}{783}\right)\)
\(B=9.\dfrac{1}{3}-9.\dfrac{1}{783}\)
\(B=3-\dfrac{9}{783}< 3\)
\(\Rightarrow B< 3\rightarrowđpcm\)
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Tính tổng các phân số sau :
a) A = \(\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+........+\dfrac{36}{45.47.49}\)
b) B= \(\left(1-\dfrac{1}{3}\right).\left(1-\dfrac{1}{9}\right).\left(1-\dfrac{1}{16}\right)......\left(1-\dfrac{1}{10000}\right)\)
c) C = 3+33+333+3333+...........+333........333
dùng công thức \(\dfrac{2m}{a\left(a+m\right)\left(a+2m\right)}=\dfrac{1}{a\left(a+m\right)}-\dfrac{1}{\left(a+m\right)\left(a+2m\right)}\)để chứng tỏ rằng:
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
\(2A=\dfrac{2}{1.2.3}+\dfrac{2}{2.3.4}+\dfrac{2}{3.4.5}+...+\dfrac{2}{18.19.20}\)
\(\Rightarrow2A=\dfrac{1}{1.2}-\dfrac{1}{2.3}+\dfrac{1}{2.3}-\dfrac{1}{3.4}+\dfrac{1}{3.4}-\dfrac{1}{4.5}+...+\dfrac{1}{18.19}-\dfrac{1}{19.20}\)
\(\Rightarrow2A=\dfrac{1}{1.2}-\dfrac{1}{19.20}< \dfrac{1}{1.2}\)
\(\Rightarrow2A< \dfrac{1}{2}\)
\(\Rightarrow A< \dfrac{1}{4}\) (đpcm)
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dùng công thức \(\dfrac{2m}{a\left(a+m\right)\left(a+2m\right)}=\dfrac{1}{a\left(a+m\right)}-\dfrac{1}{\left(a+m\right)\left(a+2m\right)}\)để chứng tỏ rằng:
\(A=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+...+\dfrac{1}{18.19.20}< \dfrac{1}{4}\)
Bài 1 : Tính giá trị chưa biết :
1^1_{30}:(24^1_6-24^1_5)-1^1_2-dfrac{3}{4}:4x-dfrac{1}{2}(-1^1_{15}):left(8^1_5-8^1_3right)
Bài 2 : Chứng Minh Rằng
A dfrac{36}{1.3.5}+dfrac{36}{3.5.7}+...+dfrac{36}{25.27.29} 3
B dfrac{1}{12}+dfrac{1}{2^2}+...+dfrac{1}{50^2} 2
Bài 3 : Cho P dfrac{2}{60.63}+dfrac{2}{63.66}+...+dfrac{2}{117+120}+dfrac{2}{2011}
Q dfrac{5}{40.44}+dfrac{5}{44.48}+...+dfrac{5}{76.80}+dfrac{5}{2011}
So sánh P và Q
Đọc tiếp
Bài 1 : Tính giá trị chưa biết :
\(1^1_{30}:(24^1_6-24^1_5)-1^1_2-\dfrac{3}{4}:4x-\dfrac{1}{2}=(-1^1_{15}):\left(8^1_5-8^1_3\right)\)
Bài 2 : Chứng Minh Rằng
A = \(\dfrac{36}{1.3.5}+\dfrac{36}{3.5.7}+...+\dfrac{36}{25.27.29}< 3\)
B= \(\dfrac{1}{12}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}< 2\)
Bài 3 : Cho P = \(\dfrac{2}{60.63}+\dfrac{2}{63.66}+...+\dfrac{2}{117+120}+\dfrac{2}{2011}\)
Q =\(\dfrac{5}{40.44}+\dfrac{5}{44.48}+...+\dfrac{5}{76.80}+\dfrac{5}{2011}\)
So sánh P và Q
2.
\(A=\dfrac{36}{1\cdot3\cdot5}+\dfrac{36}{3\cdot5\cdot7}+...+\dfrac{36}{25\cdot27\cdot29}\\ =9\cdot\left(\dfrac{4}{1\cdot3\cdot5}+\dfrac{4}{3\cdot5\cdot7}+...+\dfrac{4}{25\cdot27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{3\cdot5}+\dfrac{1}{3\cdot5}-\dfrac{1}{5\cdot7}+...+\dfrac{1}{25\cdot27}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{1\cdot3}-\dfrac{1}{27\cdot29}\right)\\ =9\cdot\left(\dfrac{1}{3}-\dfrac{1}{783}\right)\\ =9\cdot\dfrac{1}{3}-9\cdot\dfrac{1}{783}\\ =3-\dfrac{1}{87}< 3\)
Vậy \(A< 3\)
b,
\(B=\dfrac{1}{1^2}+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B=1+\dfrac{1}{2^2}+...+\dfrac{1}{50^2}\\ B< 1+\dfrac{1}{1\cdot2}+\dfrac{1}{2\cdot3}+...+\dfrac{1}{49\cdot50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+...+\dfrac{1}{49}-\dfrac{1}{50}\\ B< 1+\dfrac{1}{1}-\dfrac{1}{50}\\ B< 2-\dfrac{1}{50}< 2\)
Vậy \(B< 2\)
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\(P=\dfrac{2}{60\cdot63}+\dfrac{2}{63\cdot66}+...+\dfrac{2}{117\cdot120}+\dfrac{2}{2011}\\ =\dfrac{2}{3}\cdot\left(\dfrac{3}{60\cdot63}+\dfrac{3}{63\cdot66}+...+\dfrac{3}{117\cdot120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{63}+\dfrac{1}{63}-\dfrac{1}{66}+...+\dfrac{1}{117}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{60}-\dfrac{1}{120}+\dfrac{3}{2011}\right)\\ =\dfrac{2}{3}\cdot\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)\)
\(Q=\dfrac{5}{40\cdot44}+\dfrac{5}{44\cdot48}+...+\dfrac{5}{76\cdot80}+\dfrac{5}{2011}\\ =\dfrac{5}{4}\cdot\left(\dfrac{4}{40\cdot44}+\dfrac{4}{44\cdot48}+...+\dfrac{4}{76\cdot80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{44}+\dfrac{1}{44}-\dfrac{1}{48}+...+\dfrac{1}{76}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{40}-\dfrac{1}{80}+\dfrac{4}{2011}\right)\\ =\dfrac{5}{4}\cdot\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\)
\(\dfrac{3}{2011}< \dfrac{4}{2011}\Rightarrow\dfrac{1}{2}+\dfrac{3}{2011}< \dfrac{1}{2}+\dfrac{4}{2011}\left(1\right)\)
\(\dfrac{2}{3}< \dfrac{5}{4}\left(2\right)\)
Từ (1) và (2) ta có: \(\dfrac{2}{3}\left(\dfrac{1}{2}+\dfrac{3}{2011}\right)< \dfrac{5}{4}\left(\dfrac{1}{2}+\dfrac{4}{2011}\right)\Leftrightarrow P< Q\)
Vậy P < Q
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1.Tính :
a ) Sdfrac{1}{1.2}+dfrac{1}{2.3}+dfrac{1}{3.4}+....+dfrac{1}{2018.2019}
b ) Sdfrac{1}{1.3}+dfrac{1}{3.5}+dfrac{1}{5.7}+....+dfrac{1}{2017.2019}
c) Sdfrac{1}{2.4}+dfrac{1}{4.6}+dfrac{1}{6.8}+....+dfrac{1}{2018.2020}
d) Sdfrac{1}{1.2.3}+dfrac{1}{2.3.4}+dfrac{1}{3.4.5}+....+dfrac{1}{2017.2018.2019}
2. Tính tổng:
a) S1.2+2.3+3.4+...+2018.2019
b) S3.5+5.7+7.9+...+2017.2019
c) S2.4+4.6+6.8+...+2018.2020
d) S1.2.3+2.3.4+3.4.5+...+2017.2018.2019
3.Tính
a ) Sdfrac{1}{1.4}+dfrac{1}{4....
Đọc tiếp
1.Tính :
a ) \(S=\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+....+\dfrac{1}{2018.2019}\)
b ) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+....+\dfrac{1}{2017.2019}\)
c) \(S=\dfrac{1}{2.4}+\dfrac{1}{4.6}+\dfrac{1}{6.8}+....+\dfrac{1}{2018.2020}\)
d) \(S=\dfrac{1}{1.2.3}+\dfrac{1}{2.3.4}+\dfrac{1}{3.4.5}+....+\dfrac{1}{2017.2018.2019}\)
2. Tính tổng:
a) \(S=1.2+2.3+3.4+...+2018.2019\)
b) \(S=3.5+5.7+7.9+...+2017.2019\)
c) \(S=2.4+4.6+6.8+...+2018.2020\)
d) \(S=1.2.3+2.3.4+3.4.5+...+2017.2018.2019\)
3.Tính
a ) \(S=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+.....+\dfrac{1}{2017.2020}\)
b ) \(S=\dfrac{1}{1.3.5}+\dfrac{1}{3.5.7}+\dfrac{1}{5.7.9}+....+\dfrac{1}{2017.2019.2021}\)
Ai có công thức không cho mình xin với ????
Bài 1a) \(\dfrac{1}{1.2}+\dfrac{1}{2.3}+\dfrac{1}{3.4}+...+\dfrac{1}{2018.2019}\)
\(=1-\dfrac{1}{2}+\dfrac{1}{2}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{4}+....+\dfrac{1}{2018}-\dfrac{1}{2019}\)
\(=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
b) \(S=\dfrac{1}{1.3}+\dfrac{1}{3.5}+\dfrac{1}{5.7}+...+\dfrac{1}{2017.2019}\)
\(2S=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{2017.2019}\)
\(2S=1-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{2017}-\dfrac{1}{2019}\)
\(2S=1-\dfrac{1}{2019}=\dfrac{2018}{2019}\)
\(S=\dfrac{1009}{2019}\)
Còn lại bạn làm tương tự hết nhé .
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