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BC
31 tháng 3 2019
a)Ta có:
\(A=4\frac{25}{16}+25\left(\frac{9}{16}:\frac{125}{64}\right):\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+25.\frac{36}{125}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{36}{5}:\frac{-27}{8}\)
\(\Rightarrow A=\frac{89}{16}+\frac{-32}{15}\)
\(\Rightarrow A=\frac{823}{240}\)
Vậy A=.....
b)Ta có:
\(C=\frac{2^3}{3.5}+\frac{2^3}{5.7}+\frac{2^3}{7.9}+...+\frac{2^3}{101.103}\)
\(\Rightarrow C=\frac{2^2.2}{3.5}+\frac{2^2.2}{5.7}+\frac{2^2.2}{7.9}+...+\frac{2^2.2}{101.103}\)
\(\Rightarrow C=2^2\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+...+\frac{2}{101.103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-\frac{1}{7}+\frac{1}{7}-\frac{1}{9}+....+\frac{1}{101}-\frac{1}{103}\right)\)
\(\Rightarrow C=4\left(\frac{1}{3}-\frac{1}{103}\right)\)
\(\Rightarrow C=4.\frac{100}{309}\)
\(\Rightarrow C=\frac{400}{309}\)
Vậy C=.....
15 tháng 2 2020
A=\(\left(\frac{1}{4}-1\right).\left(\frac{1}{9}-1\right).\left(\frac{1}{16}-1\right).............\left(\frac{1}{9801}-1\right).\left(\frac{1}{10000}-1\right)\)
A=\(\left(\frac{1-4}{4}\right).\left(\frac{1-9}{9}\right).\left(\frac{1-16}{16}\right).............\left(\frac{1-9801}{9801}\right).\left(\frac{1-10000}{10000}\right)\)
A=\(\frac{-3}{4}.\frac{-8}{9}.\frac{-15}{16}.....................\frac{-9800}{9801}.\frac{-9999}{10000}\)
A=\(\frac{-1.3}{2^2}.\frac{-2.4}{3^2}.\frac{-3.5}{4^2}.....................\frac{-98.100}{99^2}.\frac{-99.101}{100^2}\)
A=\(\frac{\left[\left(-1\right).\left(-2\right).\left(-3\right)....................\left(-98\right).\left(-99\right)\right].\left(3.4.5............100.101\right)}{\left(2.3.4.........99.100\right).\left(2.3.4...............99.100\right)}\)
A=\(\frac{1.101}{100.2}\)=\(\frac{101}{200}\)
2
\(\frac{1}{3}+\frac{1}{6}+\frac{1}{10}+.................+\frac{2}{x.\left(x+1\right)}=\frac{2015}{2017}\)
\(\frac{1}{3.2}+\frac{1}{6.2}+\frac{1}{10.2}+.................+\frac{2}{2.x.\left(x+1\right)}=\frac{1}{2}.\frac{2015}{2017}\)
\(\frac{1}{6}+\frac{1}{12}+\frac{1}{20}+.................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2.3}+\frac{1}{3.4}+\frac{1}{4.5}+..................+\frac{1}{x.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{3}+\frac{1}{3}-\frac{1}{4}+\frac{1}{4}-\frac{1}{5}+..............+\frac{1}{x}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{1}{2}-\frac{1}{x+1}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x+1}{2.\left(x+1\right)}-\frac{2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{\left(x+1\right)-2}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
\(\frac{x-1}{2.\left(x+1\right)}=\frac{2015}{2017}.\frac{1}{2}\)
=>\(\frac{x-1}{x+1}=\frac{2015}{2017}.\frac{1}{2}:\frac{1}{2}\)
\(\frac{x-1}{x+1}=\frac{2015}{2017}\)
=>x+1=2017
=>x=2018-1
=>x=2016
Vậy x=2016
Còn bài 3 em ko biết làm em ms lớp 6
Chúc anh học tốt
\(\left(\frac{1}{9}\right)^{2015}.9^{2015}-96^2:24^2\)
\(=1-\left(4\right)^2\)
\(=1-16=-15\)