tính hợp lí câu sau
\(\frac{3}{4}.\frac{8}{9}.\frac{15}{16}...\frac{9999}{10000}\)
giúp mk với , mai mk đi học rồi
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\(\frac{3}{4}\)*\(\frac{8}{9}\)*\(\frac{15}{16}\)********\(\frac{9999}{10000}\)
= \(\frac{1\cdot3}{2^2}\)*\(\frac{2\cdot4}{3^2}\)********\(\frac{99\cdot101}{100^2}\)
= \(\frac{1\cdot2\cdot3\cdot4\cdot\cdot\cdot\cdot99}{2\cdot3\cdot4\cdot\cdot\cdot\cdot100}\)* \(\frac{3\cdot4\cdot5\cdot\cdot\cdot101}{2\cdot3\cdot4\cdot\cdot\cdot100}\)
= \(\frac{1}{100}\)*\(\frac{101}{2}\)=\(\frac{101}{200}\)
Ta có: A = \(\frac{3}{8}\). \(\frac{8}{9}\).\(\frac{15}{16}\). ... .\(\frac{9999}{10000}\)
\(\Rightarrow\) A = \(\frac{1.3}{2^2}\).\(\frac{2.4}{3^2}\). \(\frac{3.5}{4^2}\). ... . \(\frac{99.101}{100^2}\)
\(\Rightarrow\) A = \(\frac{1.111}{2.100}\)= \(\frac{111}{200}\)
Vậy: A = \(\frac{111}{200}\).
\(=\left(25+\dfrac{12}{67}+9+\dfrac{13}{41}-8-\dfrac{12}{67}+3+\dfrac{28}{41}\right)\cdot\dfrac{-21}{13}\)
\(=\left(25+9-8+3+1\right)\cdot\dfrac{-21}{13}=\dfrac{-630}{13}\)
\(x=\frac{3}{4}+\frac{8}{9}+\frac{15}{16}+...+\frac{9999}{10000}\)
\(x=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)
\(x=\frac{1.2...99}{2.3...100}.\frac{3.4...101}{2.3...100}\)
\(x=\frac{1}{100}.\frac{101}{2}\)
\(x=\frac{101}{200}\)
\(X=\frac{1.3}{2.2}+\frac{2.4}{3.3}+\frac{3.5}{4.4}+...+\frac{99.101}{100.100}\)
\(X=\frac{1.2.3....99}{2.3.4....100}.\frac{3.4.5....101}{2.3.4....100}\)
\(X=\frac{1}{100}.\frac{101}{2}\)
\(X=\frac{101}{200}\)
Study well
\(A=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.....\frac{9999}{10000}=\frac{3.8.15....9999}{4.9.16....10000}=?\)
P=1.3/2.2 . 2.4/3.3 . 3.5/4.4 ... . 99.101/100.100
P=1.2.3....99/2.3.4...100 . 3.4.5...101/2.3.4...100
P=1/100 . 101/2
P=101/200
A=1.3/2^2.2.4/3^2.3.5/4^2...99.101/100.100
A=(1.2.3...99/2.3.4...100).(3.4.5...101/2.3.4...100)
A=1/100.101/2
A=101/200
200.A=200.101/200
200.A=101
\(M=\frac{3}{4}.\frac{8}{9}.....\frac{9999}{10000}=\frac{1\cdot3}{2\cdot2}\cdot\frac{2\cdot4}{3\cdot3}\cdot....\cdot\frac{99\cdot101}{100\cdot100}=\frac{1\cdot3\cdot2\cdot4\cdot...\cdot99\cdot101}{2^2\cdot3^2\cdot...\cdot100^2}=\frac{1\cdot101}{2\cdot100}=\frac{101}{200}\)Vậy M = \(\frac{101}{200}\)
\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}....\frac{9999}{10000}\)
\(M=\frac{1.3}{2^2}.\frac{2.4}{3^2}.\frac{3.5}{4^2}....\frac{99.101}{100^2}=\frac{1.2.3...99}{2.3.4...100}.\frac{3.4.5...101}{2.3.4...100}=\frac{1}{100}.\frac{101}{2}=\frac{101}{200}\)
\(\frac{3}{2^2}.\frac{2^3}{3^2}.\frac{5.3}{4^2}.......\frac{3.3333}{100^2}\)
mk ko nghĩ ra phần sau chắc là rút gọn thì phải!!?
bạn làm thử xem
mk nghi ra thi da ko phai hoi