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\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{99}{100}\)
\(=\frac{3.8.15.24....99}{4.9.16.25....100}\)
\(=\frac{1.3.2.4.3.5.4.6....9.11}{2.2.3.3.4.4.5.5....10.10}\)
\(=\frac{1.2.3.4....9}{2.3.4.5....10}.\frac{3.4.5.6....11}{2.3.4.5....10}\)
\(=\frac{1}{10}.\frac{11}{2}\)
\(=\frac{11}{20}\)
Study well ! >_<
\(M=\frac{3}{4}.\frac{8}{9}.\frac{15}{16}.\frac{24}{25}...\frac{99}{100}\)
\(\Rightarrow M=\frac{1.3}{2.2}.\frac{2.4}{3.3}.\frac{3.5}{4.4}.\frac{4.6}{5.5}...\frac{9.11}{10.10}\)
\(\Rightarrow M=\frac{1.3.2.4.3.5...9.11}{2.2.3.3.4.4...10.10}\)
\(\Rightarrow M=\frac{\left(1.2.3...9\right)\left(3.4.5...11\right)}{\left(2.3.4...10\right)\left(2.3.4...10\right)}\)
\(\Rightarrow M=\frac{11}{10.2}\)
\(\Rightarrow M=\frac{11}{20}\)
a) \(\frac{8}{9}=1-\frac{1}{9}\)
\(\frac{108}{109}=1-\frac{1}{109}\)
Vì \(\frac{1}{9}>\frac{1}{109}\)
Nên \(1-\frac{1}{9}< 1-\frac{1}{109}\)
Vậy \(\frac{8}{9}< \frac{108}{109}\)
b)
\(\frac{97}{100}=\frac{97\cdot99}{100\cdot99}\)
\(\frac{98}{99}=\frac{98\cdot100}{99\cdot100}\)
\(\Rightarrow\frac{97}{100}< \frac{98}{99}\)
p) \(\frac{-5}{9}+\frac{8}{15}+\frac{-2}{11}+\frac{4}{-9}+\frac{7}{15}\)
\(=\left(\frac{-5}{9}+\frac{-4}{9}\right)+\left(\frac{8}{15}+\frac{7}{15}\right)+\frac{-2}{11}\)
\(=-1+1+\frac{-2}{11}\)
\(=-\frac{2}{11}\)
q) \(\frac{5}{13}+\frac{-5}{17}+\frac{-20}{41}+\frac{8}{13}+\frac{-21}{41}\)
\(=\left(\frac{5}{13}+\frac{8}{13}\right)+\left(\frac{-20}{41}+\frac{-21}{41}\right)+\frac{-5}{17}\)
\(=1+\left(-1\right)+\frac{-5}{17}=\frac{-5}{17}\)
r) \(\frac{1}{5}+\frac{-2}{9}+\frac{-7}{9}+\frac{4}{5}+\frac{16}{17}\)
\(=\left(\frac{1}{5}+\frac{4}{5}\right)+\left(\frac{-2}{9}+\frac{-7}{9}\right)+\frac{16}{17}\)
\(=1+\left(-1\right)+\frac{16}{17}=\frac{16}{17}\)
A = 1/3 -. 3 /4 + 3/5 + 1/64 -2/ 9 - 1 /36 +1/ 15
A = ( 1 /3 +3 /5 +1/ 15) - ( 3/4 -2/9 -1/ 36) + 1/64
A =1-1 +1/64 = 1/64
Muốn cho số có hai chữ số giống nhau và chia hết cho 2 thì số đó phải là một trong các số 22, 44, 66, 88. Bây giờ ta tìm trong những số này số mà chia cho 5 thì dư 3.
Đó là số 88.
Xem thêm tại: http://loigiaihay.com/bai-99-trang-39-sgk-toan-6-tap-1-c41a3896.html#ixzz4xczZ4dOb
ta gọi \(\frac{1}{2}+\frac{1}{6}+...+\frac{1}{90}\)là A
\(A=\frac{1}{1.2}+\frac{1}{2.3}+\frac{1}{3.4}+...+\frac{1}{9.10}\)
\(\Leftrightarrow1.\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+....+\frac{1}{9}-\frac{1}{10}\right)\)
\(\Rightarrow A=1-\frac{1}{10}=\frac{9}{10}\)
ta gọi B là biểu thức thứ2
\(B=\frac{2.2}{3}\times\frac{3.3}{2.4}\times\frac{4.4}{3.5}\times...\times\frac{10.10}{9.11}\)
\(\Rightarrow\)2 x \(\frac{10}{11}\)\(=\frac{20}{11}\)
\(\Rightarrow\)\(x+\frac{9}{10}=\frac{20}{11}+\frac{9}{110}\)
\(\Rightarrow x=1\)
mk nghĩ vậy bạn ạ, mk mong nó đúng
a) \(\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)
\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)
= \(\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)
\(=\frac{0}{5}+\frac{0}{21}+\frac{3}{20}\)
\(=0+0+\frac{3}{20}\)
\(=\frac{3}{20}\)
b) \(\frac{42}{46}+\frac{250}{186}+\frac{-2121}{2323}+\frac{-125125}{143143}\)
\(=\frac{21}{23}+\frac{125}{93}+\frac{-21}{23}+\frac{-125}{143}\)
\(=\left(\frac{21}{23}+\frac{-21}{23}\right)+\frac{125}{93}+\frac{-125}{143}\)
\(=0+\frac{125}{93}+\frac{-125}{143}\)
\(=\frac{17875}{13299}+\frac{-11625}{13299}\)
\(=\frac{6250}{13299}\)
(Hình như câu b hơi sai)
\(a)\frac{4}{20}+\frac{16}{42}+\frac{6}{15}+\frac{-3}{5}+\frac{2}{21}+\frac{10}{21}+\frac{3}{20}\)
\(=\frac{1}{5}+\frac{8}{21}+\frac{2}{5}+\frac{-3}{5}+\frac{2}{21}+\frac{-10}{21}+\frac{3}{20}\)
\(=\left(\frac{1}{5}+\frac{2}{5}+\frac{-3}{5}\right)+\left(\frac{8}{21}+\frac{2}{21}+\frac{-10}{21}\right)+\frac{3}{20}\)
\(=0+0+\frac{3}{20}\)
\(=\frac{3}{20}\)
B = 1.3/2.2 . 2.4/3.3 . 3.5/4.4 . ...... . 9.11/10.10
= 1.2.3 . ..... . 9/2.3.4 . ..... . 10 . 3.4.5 . ...... . 11/2.3.4 . ..... . 10
= 1/10 . 11/2
= 11/20
Tk mk nha
\(B=\frac{3}{4}.\frac{8}{9}...\frac{99}{100}\)
\(B=\frac{1\cdot3}{2^2}.\frac{2\cdot4}{3^3}...\frac{9\cdot11}{10^2}\)
\(B=\frac{1.3.2.4...9.11}{2^2.3^3...10^2}\)
\(B=\frac{11}{2.10}\)
\(B=\frac{11}{20}\)