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a)\(\frac{-5}{13}+\left(\frac{3}{5}+\frac{3}{13}-\frac{4}{10}\right)=\frac{-5}{13}-\frac{3}{5}-\frac{3}{13}+\frac{4}{10}=\left(\frac{-5}{13}-\frac{3}{13}\right)+\frac{4}{10}-\frac{3}{5}=\frac{-5-3}{13}+\left(\frac{4}{10}-\frac{6}{10}\right)=\frac{-8}{13}+\frac{-2}{10}=\frac{-80}{130}+\frac{-26}{130}=\frac{-106}{130}=\frac{-53}{65}\)
Bài 2: Mỗi xe ô tô có 4 bánh xe . Hỏi 5 xe ô tô như thế có bao nhiêu bánh xe ?
Bài giải
5 xe ô tô như thế có số bánh xe là :
4 x 5= 20 (bánh xe )
Đáp số : 20 bánh xe
\(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)
\(=\frac{2}{4}=\frac{1}{2}\)
\(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\)
\(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)
\(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\)
\(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\)
\(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\) \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\) \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\) \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\)
\(=\frac{58}{7}-\frac{487}{63}\) \(=\frac{577}{45}-\frac{280}{45}\)
\(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\) \(=\frac{33}{5}\)
\(P=M-N\)
\(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\)
\(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\)
\(\Rightarrow P=\frac{-272}{45}\)
Vậy P = \(\frac{-272}{45}\)
\(6)E=10101\left(\frac{5}{111111}+\frac{5}{222222}-\frac{4}{3.7.11.13.37}\right)\)
\(=\frac{5}{11}+\frac{5}{22}-\left(10101.\frac{4}{111111}\right)\)
\(=\frac{10}{22}+\frac{5}{22}-\frac{4}{11}\)
\(=\frac{15}{22}-\frac{8}{22}=\frac{7}{22}\)
\(7)F=\frac{\frac{1}{3}+\frac{1}{7}-\frac{1}{13}}{\frac{2}{3}+\frac{2}{7}-\frac{2}{13}}.\frac{\frac{3}{4}-\frac{3}{16}-\frac{3}{256}+\frac{3}{64}}{1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}}+\frac{5}{8}\)
\(=\frac{1\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}{2\left(\frac{1}{3}+\frac{1}{7}-\frac{1}{13}\right)}.\frac{3\left(\frac{1}{4}-\frac{1}{16}-\frac{1}{256}+\frac{1}{64}\right)}{1\left(1-\frac{1}{4}+\frac{1}{16}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{16}{64}-\frac{4}{64}+\frac{1}{64}-\frac{1}{256}\right)}{1\left(\frac{64}{64}-\frac{16}{64}+\frac{4}{64}-\frac{1}{64}\right)}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{13}{64}-\frac{1}{256}\right)}{1.\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{52}{256}-\frac{1}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3\left(\frac{51}{256}\right)}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{\frac{153}{256}}{\frac{51}{64}}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{153}{256}:\frac{51}{64}+\frac{5}{8}\)
\(=\frac{1}{2}.\frac{3}{4}+\frac{5}{8}\)
\(=\frac{3}{8}+\frac{5}{8}=1\)
Xin lỗi tớ đã làm hết buổi tối mà chỉ có 7 bài mong bạn thông cảm cho mình nhé !
Bài 1:
a) \(\frac{\left(-3\right)}{16}+\frac{1}{15}=\frac{-45}{240}+\frac{16}{240}\)
\(=\frac{-29}{240}\)
b)\(\frac{\left(-15\right)}{24}-\frac{\left(-2\right)}{6}=\frac{\left(-15\right)}{24}-\frac{-8}{24}\)
\(=\frac{-7}{24}\)
c) \(\frac{\left(-16\right)}{18}\cdot\frac{36}{\left(-40\right)}=\frac{\left(-8\right)}{9}\cdot\frac{\left(-9\right)}{10}\)
\(=\frac{\left(-80\right)}{90}\cdot\frac{\left(-81\right)}{90}\)
\(=\frac{4}{5}\)
d)\(\frac{\left(-17\right)}{30}:\frac{34}{60}=\frac{\left(-17\right)}{30}:\frac{17}{30}\)
\(=\frac{\left(-17\right)}{30}\cdot\frac{30}{17}\)
\(=-1\)
Bài 2:
a) \(1\frac{3}{5}+2\frac{1}{6}=\frac{8}{5}+\frac{13}{6}=\frac{48}{30}+\frac{65}{30}\)
\(=\frac{113}{30}\)
b) \(3\frac{1}{7}-1\frac{1}{8}=\frac{22}{7}-\frac{9}{8}=\frac{176}{56}-\frac{63}{56}\)
\(=\frac{113}{56}\)
c) \(3\frac{1}{6}\cdot2\frac{1}{4}=\frac{19}{6}\cdot\frac{9}{4}=\frac{57}{8}\)
d) \(4\frac{1}{5}:3\frac{6}{7}=\frac{21}{5}:\frac{27}{7}=\frac{21}{5}\cdot\frac{7}{27}\)
\(=\frac{49}{45}\)
a \(\frac{-3}{16}+\frac{1}{15}=\frac{-45}{240}+\frac{16}{240}=\frac{-29}{240}\)
b \(\frac{-15}{24}-\frac{-2}{6}=\frac{-15}{24}-\frac{-8}{24}=\frac{-7}{24}\)
c \(\frac{-16}{18}.\frac{36}{-40}=\frac{4}{5}\)
d \(\frac{-17}{30}:\frac{34}{60}=\frac{-17}{30}.\frac{60}{34}=-1\)
bai 2
\(1\frac{3}{5}+2\frac{1}{6}=\frac{8}{5}+\frac{13}{6}=\frac{113}{30}\)
\(3\frac{1}{7}-1\frac{1}{8}=\frac{22}{7}-\frac{9}{8}=\frac{113}{56}\)
c \(3\frac{1}{6}.2\frac{1}{4}=\frac{19}{6}.\frac{9}{4}=\frac{57}{8}\)
d \(4\frac{1}{5}:3\frac{6}{7}=\frac{21}{5}:\frac{27}{7}=\frac{21}{5}.\frac{7}{27}=\frac{147}{135}\)