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31 tháng 7 2019
\(B=\frac{1}{3}-\frac{3}{4}+0,6+\frac{1}{64}-\frac{2}{9}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}-\frac{48}{64}+\frac{9}{15}+\frac{1}{64}-\frac{8}{36}-\frac{1}{36}+\frac{1}{15}\)
\(\Rightarrow B=\frac{3}{15}+\frac{9}{15}+\frac{1}{15}+\left(-\frac{48}{64}+\frac{1}{64}\right)+\left(-\frac{8}{36}-\frac{1}{36}\right)\)
\(\Rightarrow B=\frac{13}{15}-\frac{47}{64}-\frac{1}{4}\)
\(\Rightarrow B=-\frac{113}{960}\)
\(C=0\)
\(D=\frac{1}{99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)
\(\Rightarrow D=\frac{1}{99}-\frac{1}{99}+\frac{1}{98}-\frac{1}{98}+...-\frac{1}{3}+\frac{1}{2}-\frac{1}{2}+1\)
\(\Rightarrow D=1\)
LT
11 tháng 8 2019
D= \(\frac{1}{99}-\frac{1}{99.98}-\frac{1}{98.97}......-\frac{1}{3.2}-\frac{1}{2.1}\)
=\(\frac{1}{99}-\left(\frac{1}{1.2}+\frac{1}{2.3}+.......+\frac{1}{97.98}+\frac{1}{98.99}\right)\)
=\(\frac{1}{99}-\left(1-\frac{1}{2}+\frac{1}{2}-.....-\frac{1}{98}-\frac{1}{99}\right)\)
=\(\frac{1}{99}-\left[1-(\frac{1}{2}-\frac{1}{2}+......+\frac{1}{98}-\frac{1}{99})\right]\)
=\(\frac{1}{99}-\left(1-0-0-.....-0-\frac{1}{99}\right)\)
=\(\frac{1}{99}-1-\frac{1}{99}\)
=1
VM
17 tháng 11 2019
Ta có:
\(0,4\left(3\right)=\frac{43-4}{90}=\frac{39}{90}=\frac{13}{30}.\)
\(0,6\left(2\right)=\frac{62-6}{90}=\frac{56}{90}=\frac{28}{45}.\)
\(0,6\left(8\right)=\frac{68-6}{90}=\frac{62}{90}=\frac{31}{45}.\)
Vậy:
\(\frac{13}{30}+\frac{28}{45}.\frac{5}{2}-\frac{\frac{5}{6}}{\frac{31}{45}}.\frac{53}{50}\)
\(=\frac{13}{30}+\frac{14}{9}-\frac{75}{62}.\frac{53}{50}\)
\(=\frac{13}{30}+\frac{14}{9}-\frac{159}{124}\)
\(=\frac{179}{90}-\frac{159}{124}\)
\(=\frac{3943}{5580}.\)
Chúc bạn học tốt!
MA
31 tháng 8 2016
\(B=\frac{0,6-\frac{3}{11}+\frac{3}{13}}{1,4-\frac{7}{11}+\frac{7}{13}}-\frac{\frac{1}{3}-0,25+\frac{1}{5}}{1\frac{1}{6}-0,875+0,7}\)
\(B=\frac{\frac{3}{5}-\frac{3}{11}+\frac{3}{13}}{\frac{7}{5}-\frac{7}{11}+\frac{7}{13}}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{\frac{7}{6}-\frac{7}{8}+\frac{7}{10}}\)
\(B=\frac{3\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}{5\left(\frac{1}{5}-\frac{1}{11}+\frac{1}{13}\right)}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7\left(\frac{1}{6}-\frac{1}{8}+\frac{1}{10}\right)}\)
\(B=\frac{3}{5}-\frac{\frac{1}{3}-\frac{1}{4}+\frac{1}{5}}{7.\frac{1}{2}\left(\frac{1}{3}-\frac{1}{4}+\frac{1}{5}\right)}\)
\(B=\frac{3}{5}-\frac{2}{7}=\frac{11}{35}\)
HQ
Hà Quang Minh
Giáo viên
19 tháng 9 2023
a)
\(\begin{array}{l}0,6 + \left( {\frac{3}{{ - 4}}} \right) = \frac{6}{{10}} + \left( {\frac{{ - 3}}{4}} \right)\\ = \frac{{12}}{{20}} + \left( {\frac{{ - 15}}{{20}}} \right) = \frac{{12 + \left( { - 15} \right)}}{{20}}\\ = \frac{{ - 3}}{{20}}\end{array}\)
b)
\(\begin{array}{l}\left( { - 1\frac{1}{3}} \right) - \left( { - 0,8} \right) = \frac{{ - 4}}{3} + \frac{8}{{10}}\\ = \frac{{ - 4}}{3} + \frac{4}{5} = \frac{{ - 20}}{{15}} + \frac{{12}}{{15}} = \frac{{ - 8}}{{15}}.\end{array}\)
19 tháng 9 2023
a: =0,6-0,75=-0,15
b: \(=-\dfrac{4}{3}+\dfrac{4}{5}=\dfrac{-20+12}{15}=-\dfrac{8}{15}\)
QT
So sánh:
a) \( - \frac{1}{3}\) và \(\frac{{ - 2}}{5}\)
b) 0,125 và 0,13
c) -0,6 và \(\frac{{ - 2}}{3}\)
1
HQ
Hà Quang Minh
Giáo viên
16 tháng 9 2023
a) Ta có:
\( - \frac{1}{3} = \frac{{ - 5}}{{15}};\frac{{ - 2}}{5} = \frac{{ - 6}}{{15}}\)
Vì -5 > -6 nên \(\frac{{ - 5}}{{15}} > \frac{{ - 6}}{{15}}\) hay \( - \frac{1}{3}\) > \(\frac{{ - 2}}{5}\)
b) 0,125 < 0,13 vì chữ số hàng phần trăm của 0,125 là 2 nhỏ hơn chữ số hàng phần trăm của 0,13 là 3
c) Ta có:
\(\begin{array}{l} - 0,6 = \frac{{ - 6}}{{10}} = \frac{{ - 3}}{5} = \frac{{ - 9}}{{15}};\\\frac{{ - 2}}{3} = \frac{{ - 10}}{{15}}\end{array}\)
Vì -9 > -10 nên \(\frac{{ - 9}}{{15}} > \frac{{ - 10}}{{15}}\) hay - 0,6 > \(\frac{{ - 2}}{3}\)
3 tháng 8 2015
\(P=\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)
\(P=\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{5}+\frac{11}{7}+\frac{11}{13}}\)
\(P=\frac{3\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{11\cdot\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}=\frac{3}{11}\)
\(B=\frac{1}{3}+\frac{1}{3^2}+\frac{1}{3^3}+...+\frac{1}{3^{2015}}+\frac{1}{3^{2016}}\)
\(\frac{1}{B}=3+3^2+3^3+...+3^{2015}+3^{2016}\)
\(\frac{3}{B}=3^2+3^3+3^4+...+3^{2016}+3^{2017}\)
\(\frac{3}{B}-\frac{1}{B}=\left(3^2+3^3+3^4+...+3^{2016}+3^{2017}\right)-\left(3+3^2+3^3+...+3^{2015}+3^{2016}\right)\)
\(\frac{2}{B}=3^{2017}-3\)
\(B=\frac{2}{3^{2017}-3}\)
3 tháng 8 2015
P=\(\frac{0,75-0,6+\frac{3}{7}+\frac{3}{13}}{2,75-2,2+\frac{11}{7}+\frac{11}{13}}\)
P=\(\frac{\frac{3}{4}-\frac{3}{5}+\frac{3}{7}+\frac{3}{13}}{\frac{11}{4}-\frac{11}{3}+\frac{11}{7}+\frac{11}{3}}\)
P=\(\frac{\frac{1}{3}.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}{\frac{1}{11}.\left(\frac{1}{4}-\frac{1}{5}+\frac{1}{7}+\frac{1}{13}\right)}\)
P=\(\frac{\frac{1}{3}}{\frac{1}{11}}=\frac{1}{3}:\frac{1}{11}=\frac{11}{3}\)
B=\(\frac{1}{3}+\frac{1}{^{3^2}}+\frac{1}{3^3}+................+\frac{1}{3^{2015}}+\frac{1}{3^{2016}}\)
B=\(\left(\frac{1}{3}\right)^1+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{2015}+\left(\frac{1}{3}\right)^{2016}\)
2B=\(\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+\left(\frac{1}{3}\right)^4+...+\left(\frac{1}{3}\right)^{2016}+\left(\frac{1}{3}\right)^{2017}\)
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B=\(\left(\frac{1}{3}\right)^1+\left(\frac{1}{3}\right)^2+\left(\frac{1}{3}\right)^3+...+\left(\frac{1}{3}\right)^{2015}+\left(\frac{1}{3}\right)^{2016}\)
B=\(\left(\frac{1}{3}\right)^1-\left(\frac{1}{3}\right)^{2017}\)
(??…………………