Tính giá trị của A biết:
1/ \(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{9999}{10000}\)
2/ \(A=\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{3599}{3600}\)
mọi nguời cứ tính đi đuợc câu nào tôi cũng tick cho
\(B=\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot...\cdot\frac{3599}{3600}\)
B = \(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{3599}{3600}\)
= \(\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{59.61}{60.60}\)
= \(\frac{2.4.3.5.4.6...59.61}{3.3.4.4.5.5...60.60}\)
= \(\frac{2.3....59}{3.4...60}.\frac{4.5...61}{3.4...60}\)
= \(\frac{2}{60}.\frac{61}{3}\)= \(\frac{61}{90}\)
Chúc bạn học tốt!
B=\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{9999}{10000}=?\)
\(B=\frac{\left(1.3\right).\left(2.4\right).\left(3.5\right).\left(4.6\right)...\left(99.101\right)}{2^2.3^2.4^2.5^2...100^2}=\frac{\left(1.2.3.4...99\right).\left(3.4.5.6...101\right)}{\left(2.3.4.5...100\right)\left(2.3.4.5...100\right)}=\frac{1.101}{100.2}=\frac{101}{200}\)
B = \(\frac{1.3}{2^2}.\frac{2.4}{3^2}\frac{3.5}{4^2}\frac{4.6}{5^2}...\frac{99.101}{100^2}=\frac{1.3.2.4.3.5.4.6...99.101}{2.2.3.3.4.4.5.5...100.100}\)
=\(\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
Vật B = \(\frac{101}{200}\)
đúng cái đi
Cho \(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+\frac{24}{25}+...+\frac{9999}{10000}\). Chứng minh rằng : S\(\notin\)N.
Ta có : \(S=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(=\left(1-\frac{1}{4}\right)+\left(1-\frac{1}{9}\right)+\left(1-\frac{1}{16}\right)+...+\left(1-\frac{1}{10000}\right)\)
\(=\left(1+1+1+...+1\right)-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{15}+...+\frac{1}{10000}\right)\)
\(=99-\left(\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\right)< 99\)
\(\Rightarrow\)S<99 (1)
Đặt \(A=\frac{1}{4}+\frac{1}{9}+\frac{1}{16}+...+\frac{1}{10000}\)
\(=\frac{1}{2^2}+\frac{1}{3^2}+\frac{1}{4^2}+...+\frac{1}{100^2}\)
Ta có : \(\frac{1}{2^2}=\frac{1}{2.2}< \frac{1}{1.2}\)
\(\frac{1}{3^2}=\frac{1}{3.3}< \frac{1}{2.3}\)
\(\frac{1}{4^2}=\frac{1}{4.4}< \frac{1}{3.4}\)
...
\(\frac{1}{100^2}=\frac{1}{100.100}< \frac{1}{99.100}\)
\(\Rightarrow A< \frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{99.100}\)
\(A< 1-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+...+\frac{1}{99}-\frac{1}{100}\)
\(A< 1-\frac{1}{100}< 1\)
\(\Rightarrow\)S>99-1=98 (2)
Từ (1) và (2)
\(\Rightarrow\)98<S<99
\(\Rightarrow\)S\(\notin\)N
Vậy S\(\notin\)N.
\(B=\frac{8}{9}\cdot\frac{15}{16}\cdot\frac{24}{25}\cdot...\cdot\frac{3599}{3600}\)
Nhờ các bạn tính giùm.
\(B=\frac{8}{9}^.\frac{15}{16}^.\frac{24}{25}^........^.\frac{3599}{3600}\)
\(B=\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}.....\frac{59.61}{60.60}\)
B = \(\left(\frac{2.3.4.....59}{3.4.5.....60}\right).\left(\frac{4.5.6.....61}{3.4.5.....60}\right)\)
\(B=\frac{2}{60}.\frac{61}{3}\)
B = \(\frac{61}{90}\)
B= \(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{3599}{3600}\)
B= \(\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{59.61}{60.60}\)
B=\(\frac{2.61}{3.60}\)
B= \(\frac{61}{90}\)
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}=\frac{3.8.15....9999}{4.9.16....10000}=?\)
\(\frac{3}{4}x\frac{8}{9}x\frac{15}{16}x\frac{24}{25}x\frac{35}{36}x\frac{48}{49}x\frac{63}{64}\)
Tính giá trị biểu thức
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{63}{64}\)
\(=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{7.9}{8.8}\)
\(=\frac{1.3.2.4.3.5.4.6...7.9}{2.2.3.3.4.4.5.5...8.8}\)
\(=\frac{1.9}{2.8}=\frac{9}{16}\)
Bài 3 : Tính
a) A = \(\frac{3}{4}\) .\(\frac{8}{9}\) . \(\frac{15}{16}\) . ... . \(\frac{9999}{10000}\)
b) B = \(\frac{8}{9}\) . \(\frac{15}{16}\) . \(\frac{24}{25}\) . ... . \(\frac{3599}{3600}\)
Bài 4 :
Cho : A = \(\frac{1}{2}\).\(\frac{3}{4}\) . \(\frac{5}{6}\) . ... . \(\frac{9999}{10000}\)
B = \(\frac{1}{100}\)
So sánh A và B
Bài 3 : Tính
a) A = \(\frac{3}{4}\) .\(\frac{8}{9}\) . \(\frac{15}{16}\) . ... . \(\frac{9999}{10000}\)
b) B = \(\frac{8}{9}\) . \(\frac{15}{16}\) . \(\frac{24}{25}\) . ... . \(\frac{3599}{3600}\)
Bài 4 :
Cho : A = \(\frac{1}{2}\).\(\frac{3}{4}\) . \(\frac{5}{6}\) . ... . \(\frac{9999}{10000}\)
B = \(\frac{1}{100}\)
So sánh A và B
tích mình đi
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tích lại
thanks
a)\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}=\)\(\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.....\frac{99.101}{100.100}=\frac{101}{2.100}=\frac{101}{200}\)
b)\(\frac{8}{9}.\frac{15}{16}.\frac{24}{25}.....\frac{3599}{3600}=\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.6}.....\frac{59.61}{60.60}=\frac{2.61}{60}=\frac{61}{30}\)
Tính
A=\(\frac{8}{9}+\frac{15}{16}+\frac{24}{25}+...+\frac{2499}{2500}\)
a=8/9+15/16+24/25+....+2499/2500
a=(1-1/9)+(1-1/16)+(1-1/25)+....+(1-1/2500)
a=1-1/9+1-1/16+1-1/25+....+1-1/2500
a=(1+1+...+1)-(1/9+1/16+1/25+....+1/2500)