Cho A = \(\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times.......\times\frac{9999}{10000}\). So sánh A với 0.01
Những câu hỏi liên quan
A= \(\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times...\times\frac{9999}{10000}\). Hãy so sánh A và 0,01
Xem thêm câu trả lời
tính các tích sauafrac{3}{4}timesfrac{8}{9}timesfrac{15}{16}times...timesfrac{9999}{10000}bleft(1-frac{1}{4}right)timesleft(1-frac{1}{9}right)times...timesleft(1-frac{1}{10000}right)cleft(1-frac{1}{2}right)timesleft(1-frac{1}{3}right)times...timesleft(1-frac{1}{1994}right)dleft(1+frac{1}{1times3}right)timesleft(1+frac{1}{2times4}right)timesleft(1+frac{1}{3times5}right)times...timesleft(1+frac{1}{99times100}right)
Đọc tiếp
tính các tích sau
\(a=\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\times...\times\frac{9999}{10000}\)
\(b=\left(1-\frac{1}{4}\right)\times\left(1-\frac{1}{9}\right)\times...\times\left(1-\frac{1}{10000}\right)\)
\(c=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)
\(d=\left(1+\frac{1}{1\times3}\right)\times\left(1+\frac{1}{2\times4}\right)\times\left(1+\frac{1}{3\times5}\right)\times...\times\left(1+\frac{1}{99\times100}\right)\)
\(d=\left(1+\frac{1}{1.3}\right)\left(1+\frac{1}{2.4}\right)\left(1+\frac{1}{3.5}\right).........\left(1+\frac{1}{99.101}\right)\)
\(=\frac{4}{3}.\frac{9}{2.4}.............\frac{10000}{99.101}\)
\(=\frac{2.2}{3}.\frac{3.3}{2.4}.\frac{4.4}{3.5}............\frac{100.100}{99.101}\)
\(=\frac{2.3.4..........100}{2.3.4............99}.\frac{2.3.4...........100}{3.4...........101}\)
\(=100.\frac{2}{101}\)\(=\frac{200}{101}\)
Đúng 0
Bình luận (0)
\(C=\left(1-\frac{1}{2}\right)\times\left(1-\frac{1}{3}\right)\times...\times\left(1-\frac{1}{1994}\right)\)
\(=\frac{1}{2}\times\frac{2}{3}\times\frac{3}{4}\times...\times\frac{1993}{1994}\)
\(=\frac{1\times2\times3\times...\times1993}{2\times3\times4\times...\times1994}\)
\(=\frac{1}{1994}\) (Giản ước còn lại như này)
Đúng 0
Bình luận (0)
cho A=\(\frac{3}{4}\times\frac{8}{9}\times\frac{15}{16}\) .......\(\times\)\(\frac{9999}{10000}\).khi do 200\(\times\)A=...........
Cho \(A=\left(\frac{1}{2^2}-1\right)\times\left(\frac{1}{3^2}-1\right)\times\left(\frac{1}{4^2}-1\right)\times...\times\left(\frac{1}{100^2}-1\right)\)
So sánh A với \(-\frac{1}{2}\)
Cho biểu thức A= \(\frac{2}{1}\times\frac{4}{3}\times\frac{6}{5}\times\frac{8}{7}\times\frac{10}{9}\times...\times\frac{100}{99}\)Chứng minh rằng 12<A<13
1/tìm STN nhỏ nhất chia cho 5 dư 1,chia7 dư 5 2/CMR:1-frac{1}{2}+frac{1}{3}-frac{1}{4}+....+frac{1}{199}-frac{1}{200}frac{1}{101}+frac{1}{102}+...+frac{1}{200} 3/CMR:frac{51}{2}timesfrac{52}{2}times...timesfrac{100}{2}1times3times5times...times97times99 4/cho Afrac{1}{2}timesfrac{3}{4}timesfrac{5}{6}times...timesfrac{9999}{...
Đọc tiếp
1/tìm STN nhỏ nhất chia cho 5 dư 1,chia7 dư 5 2/CMR:\(1-\frac{1}{2}+\frac{1}{3}-\frac{1}{4}+....+\frac{1}{199}-\frac{1}{200}=\frac{1}{101}+\frac{1}{102}+...+\frac{1}{200}\) 3/CMR:\(\frac{51}{2}\times\frac{52}{2}\times...\times\frac{100}{2}=1\times3\times5\times...\times97\times99\) 4/cho A=\(\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times...\times\frac{9999}{10000}\) so sánh A với 0,01 5/CMR:\(\left(1+2+3+...+n\right)-7\) chia hết cho 10
tính bằng cách thuận tiện
a. \(\frac{1}{2}\times\frac{2}{3}\div\frac{4}{3}\times\frac{4}{5}\div\frac{6}{5}\times\frac{6}{7}\div\frac{7}{8}\times\frac{8}{9}\div\frac{10}{9}\)
b.\(\frac{27}{49}\times\frac{49}{50}\times\frac{15}{51}\times(\frac{5}{10}-\frac{1}{2})\)
(7x6=5+2+6x7) =
Cho : A frac{1}{4}timesfrac{3}{6}timesfrac{5}{8}times.......timesfrac{43}{46} Bfrac{3}{5}timesfrac{4}{7}times..........timesfrac{44}{47}a) so sánh A với Bb) chứng minh Afrac{1}{133}
Đọc tiếp
Cho : A = \(\frac{1}{4}\)\(\times\)\(\frac{3}{6}\)\(\times\)\(\frac{5}{8}\)\(\times\).......\(\times\)\(\frac{43}{46}\)
B=\(\frac{3}{5}\)\(\times\)\(\frac{4}{7}\)\(\times\)..........\(\times\)\(\frac{44}{47}\)
a) so sánh A với B
b) chứng minh A<\(\frac{1}{133}\)
Cho \(A=\frac{1}{2}\times\frac{3}{4}\times\frac{5}{6}\times...\times\frac{199}{200}\)và chứng minh \(A^2< \frac{1}{201}\)
ta có 1/2<2/3 ; 3/4<4/5;5/6<6/7;...;199/200<200/201
suy ra A^2=1/2^2*3/4^2*5/6^2*...*199/200^2<1/2*2/3*3/4*4/5*5/6*6/7*...*199/200/200/201
suy ra A^2<1/201(đpcm)
Đúng 0
Bình luận (0)
Ta có:
\(\frac{1}{2}< \frac{2}{3};\frac{3}{4}< \frac{4}{5};\frac{5}{6}< \frac{6}{7};...;\frac{199}{200}< \frac{200}{201}\)
\(\Rightarrow\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)
\(\Rightarrow A< \frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\)
\(\Rightarrow A^2< \left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}.....\frac{200}{201}\right)\left(\frac{1}{2}.\frac{3}{4}.\frac{5}{6}.....\frac{199}{200}\right)\)
\(\Rightarrow A^2< \frac{1}{201}\left(đpcm\right)\)
Đúng 0
Bình luận (0)
\(A=\frac{1}{2}.\frac{3}{4}.\frac{5}{6}...\frac{199}{200}\)
\(\Rightarrow A< \frac{2}{3}.\frac{4}{5}\frac{6}{7}...\frac{200}{201}\)
\(\Rightarrow A.A< A.\left(\frac{2}{3}.\frac{4}{5}.\frac{6}{7}...\frac{200}{201}\right)\)
\(\Rightarrow A^2< \frac{1}{201}\)(làm phần trc như Sakuraba Laura nhá)
Đúng 0
Bình luận (0)