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HT
0
16 tháng 10 2023
\(A=\dfrac{9}{8}-\dfrac{8}{9}+\dfrac{3}{24}+\dfrac{1}{4}-\dfrac{5}{16}+\dfrac{19}{25}-\dfrac{1}{9}+\dfrac{2}{25}-\dfrac{1}{81}\)
\(=\dfrac{9}{8}+\dfrac{1}{4}-\dfrac{5}{16}+\dfrac{1}{8}-\dfrac{8}{9}-\dfrac{1}{9}-\dfrac{1}{81}+\dfrac{19}{25}+\dfrac{2}{25}\)
\(=\dfrac{10}{8}+\dfrac{1}{4}-\dfrac{5}{16}-1-\dfrac{1}{81}+\dfrac{21}{25}\)
\(=\dfrac{20+4-5}{16}-\dfrac{82}{81}+\dfrac{21}{25}\)
\(=\dfrac{19}{16}-\dfrac{82}{81}+\dfrac{21}{25}\)
\(=\dfrac{32891}{16\cdot81\cdot25}\)
b: \(B=-\dfrac{1}{3}-\dfrac{8}{35}-\dfrac{2}{9}-\dfrac{1}{35}+\dfrac{4}{5}-\dfrac{4}{9}+\dfrac{3}{7}\)
\(=\dfrac{-1}{3}-\dfrac{2}{9}-\dfrac{4}{9}-\dfrac{8}{35}-\dfrac{1}{35}+\dfrac{4}{5}+\dfrac{3}{7}\)
\(=\dfrac{-3-2-4}{9}+\dfrac{-9}{35}+\dfrac{28+15}{35}\)
\(=-1+\dfrac{-9+43}{35}=-1+\dfrac{34}{35}=-\dfrac{1}{35}\)
12 tháng 11 2021
c: \(=\dfrac{3}{2}\cdot1-1-20=\dfrac{3}{2}-21=\dfrac{-39}{2}\)
JN
6 tháng 8 2018
a , \(\frac{7}{8}:\frac{1}{6}+\frac{7}{8}.\frac{-7}{18}\)
= \(\frac{21}{4}+\frac{-49}{144}=\frac{707}{144}\)
b, -1 : (-5) + \(\frac{1}{15}-\frac{-1}{15}\)
= \(\frac{1}{5}+0=\frac{1}{5}\)
c, \(\frac{9}{10}-\frac{1}{10.9}-\frac{1}{9.8}-\frac{1}{8.7}-\frac{1}{7.6}-\frac{1}{6.5}-\frac{1}{5.4}-\frac{1}{4.3}-\frac{1}{3.2}-\frac{1}{2.1}\)
= \(\frac{9}{10}-\frac{10-9}{10.9}-\frac{9-8}{9.8}-\frac{8-7}{8.7}-\frac{7-6}{7.6}-\frac{6-5}{6.5}-\frac{5-4}{5.4}-\frac{4-3}{4.3}-\frac{3-2}{3.2}.\frac{2-1}{2.1}\)
= \(\frac{9}{10}-1-\frac{1}{10}-1-\frac{1}{9}-1-\frac{1}{8}-1-\frac{1}{7}-1-\frac{1}{6}-1-\frac{1}{5}-1-\frac{1}{4}-1-\frac{1}{3}-1-\frac{1}{2}\)
= \(\frac{9}{10}-\left(1+1+1+1+1+1+1+1+1\right)-\left(\frac{1}{10}+\frac{1}{9}+\frac{1}{8}+...+\frac{1}{2}\right)\)
= \(\frac{9}{10}-9-1,928=\frac{9}{10}-7,071=-6.171\)
HH
10 tháng 6 2018
Bài 2:
\(A=\dfrac{2}{1.3}+\dfrac{2}{3.5}+\dfrac{2}{5.7}+...+\dfrac{2}{99.101}\)
\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{3}+\dfrac{1}{3}-\dfrac{1}{5}+\dfrac{1}{5}-\dfrac{1}{7}+...+\dfrac{1}{99}-\dfrac{1}{101}\)
\(\Leftrightarrow A=\dfrac{1}{1}-\dfrac{1}{101}\)
\(\Leftrightarrow A=\dfrac{100}{101}\)
Vậy ...
\(B=\dfrac{1}{1.4}+\dfrac{1}{4.7}+\dfrac{1}{7.10}+\dfrac{1}{10.13}+\dfrac{1}{13.16}\)
\(\Leftrightarrow B=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{4}+\dfrac{1}{4}-\dfrac{1}{7}+\dfrac{1}{7}-\dfrac{1}{10}+\dfrac{1}{10}-\dfrac{1}{13}+\dfrac{1}{13}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow B=\dfrac{1}{3}\left(\dfrac{1}{1}-\dfrac{1}{16}\right)\)
\(\Leftrightarrow B=\dfrac{1}{3}.\dfrac{15}{16}\)
\(\Leftrightarrow B=\dfrac{5}{16}\)
Vậy ...
10 tháng 6 2018
Bài 1:
B=\(\dfrac{\left(1+\dfrac{1}{2}+\dfrac{1}{4}+\dfrac{1}{8}+\dfrac{1}{16}\right)}{\left(1-\dfrac{1}{2}+\dfrac{1}{4}-\dfrac{1}{8}+\dfrac{1}{16}\right)}\)
\(=\dfrac{\left(1+\dfrac{1}{2}+\dfrac{1}{2^2}+\dfrac{1}{2^3}+\dfrac{1}{2^4}\right)}{\left(1-\dfrac{1}{2}+\dfrac{1}{2^2}-\dfrac{1}{2^3}+\dfrac{1}{2^4}\right)}\)
\(=\dfrac{\left(\dfrac{2^4+2^3+2^2+2+1}{2^4}\right)}{\left(2^4-2^3+2^2-2+1\right)}\)
\(=\dfrac{\left(2^3+2\right)\left(2+1\right)+1}{2^4}.\dfrac{2^4}{\left(2^3+2\right)\left(2-1\right)}\)
\(=\dfrac{2\left(2^2+1\right)\left(2+1\right)+1}{2\left(2^2+1\right)\left(2-1\right)+1}\)
\(=\dfrac{2.5.3+1}{2.5.1+1}\)
\(=\dfrac{31}{11}\)
\(=2,\left(81\right)\)
H
3
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
29 tháng 7 2023
\(a,\\ \left(x+\dfrac{1}{2}\right)^3=\dfrac{8}{125}=\dfrac{2^3}{5^3}\\ \left(x+\dfrac{1}{2}\right)^3=\left(\dfrac{2}{5}\right)^3\\ \Rightarrow\left(x+\dfrac{1}{2}\right)=\dfrac{2}{5}\\ x=\dfrac{2}{5}-\dfrac{1}{2}\\ x=-\dfrac{1}{10}\)
GN
GV Nguyễn Trần Thành Đạt
Giáo viên
29 tháng 7 2023
\(b,3\left|x\right|-27=\dfrac{1}{5}\\ 3\left|x\right|=\dfrac{1}{5}+27\\ 3\left|x\right|=\dfrac{136}{5}\\ \left|x\right|=\dfrac{136}{5}:3\\ \left|x\right|=\dfrac{136}{15}\\ Vậy:x=\dfrac{136}{15}.or.x=-\dfrac{136}{15}\)
\(\frac{1}{16}^3:\frac{1}{8}^2=\frac{1}{4096}:\frac{1}{64}=\frac{1}{4096}.64=\frac{1}{64}\)
Bài làm:
Ta có: \(\left(\frac{1}{16}\right)^3\div\left(\frac{1}{8}\right)^2\)
\(=\left(\frac{1}{2^4}\right)^3\div\left(\frac{1}{2^3}\right)^2\)
\(=\frac{1}{2^{12}}\div\frac{1}{2^6}\)
\(=\frac{1}{2^6}\)