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\(\frac{1}{2}.\left(\frac{4}{3}+\frac{2}{5}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{16}{3}\right)\)
\(=\frac{1}{2}.\left(\frac{20}{15}+\frac{6}{15}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{48}{9}\right)\)
\(=\frac{1}{2}.\frac{26}{15}-\frac{3}{4}.\frac{56}{9}\)
\(=\frac{13}{15}-\frac{14}{3}\)
\(=-\frac{19}{5}\)
\(\frac{1}{2}.\left(\frac{4}{3}+\frac{2}{5}\right)-\frac{3}{4}.\left(\frac{8}{9}+\frac{16}{3}\right)\)
\(=\left(\frac{1}{2}.\frac{4}{3}+\frac{1}{2}.\frac{2}{5}\right)-\left(\frac{3}{4}.\frac{8}{9}+\frac{3}{4}.\frac{16}{3}\right)\)
\(=\left(\frac{2}{3}+\frac{1}{5}\right)-\left(\frac{2}{3}+4\right)\)
\(=\frac{2}{3}+\frac{1}{5}-\frac{2}{3}-4\)
\(=\frac{1}{5}-4\)
\(=\frac{1}{5}-\frac{20}{5}=\frac{-19}{5}\)
\(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)
\(=\frac{2}{4}=\frac{1}{2}\)
\(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\)
\(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)
\(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\)
\(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\)
\(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\) \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\) \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\) \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\)
\(=\frac{58}{7}-\frac{487}{63}\) \(=\frac{577}{45}-\frac{280}{45}\)
\(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\) \(=\frac{33}{5}\)
\(P=M-N\)
\(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\)
\(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\)
\(\Rightarrow P=\frac{-272}{45}\)
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 !
1/
\(\frac{2n+1}{n-3}+\frac{3n-5}{n-3}-\frac{4n-5}{n-3}=\frac{2n+1+\left(3n-5\right)-\left(4n-5\right)}{n-3}=\frac{2n+1+3n-5-4n+5}{n-3}=\frac{n+1}{n-3}=\frac{n-3+4}{n-3}=\frac{n-3}{n-3}+\frac{4}{n-3}=1+\frac{4}{n-3}\)
Để S là số nguyên <=> n - 3 thuộc Ư(4) = {1;-1;2;-2;4;-4}
| n-3 | 1 | -1 | 2 | -2 | 4 | -4 |
| n | 4 | 2 | 5 | 1 | 7 | -1 |
Vậy...
\(\frac37+\frac{5}{13}+\frac{4}{13}\)
=\(\frac37+\left(\frac{5}{13}+\frac{4}{13}\right)\)
=\(\frac37+\frac{9}{13}\)
=\(\frac{39}{91}+\frac{63}{91}\)
=\(\frac{102}{91}\)
\(\left(\frac38+\frac{-3}{4}+\frac{7}{12}\right):\frac56+\frac12\)
=\(\left(\frac{9}{24}+\frac{-18}{24}+\frac{7}{24}\right)\times\frac65+\frac12\)
=\(\frac{5}{24}\times\frac65+\frac12\)
=\(\frac14+\frac12\)
=\(\frac14+\frac24\)
=\(\frac34\)
a) \(\frac{\left(3.4.2^{16}\right)^2}{11.2^{13}.4^{11}-16^9}=\frac{\left(3.2^2.2^{16}\right)^2}{11.2^{13}.\left(2^2\right)^{11}-\left(2^4\right)^9}\)
\(=\frac{\left(3.2^{18}\right)^2}{11.2^{13}.2^{22}-2^{36}}\)
\(=\frac{3^2.2^{36}}{11.2^{35}-2^{36}}\)
\(=\frac{9.2^{36}}{2^{35}.\left(11-2\right)}\)
\(=\frac{9.2^{36}}{2^{35}.9}=2\)
b) \(\frac{3}{2}.x-\left(\frac{4}{5}-2.x\right)=1\frac{3}{10}:\frac{3}{2}\)
\(\frac{3}{2}.x-\frac{4}{5}+2.x=\frac{13}{10}:\frac{3}{2}\)
\(\left(\frac{3}{2}.x+2.x\right)-\frac{4}{5}=\frac{13}{10}.\frac{2}{3}\)
\(x.\left(\frac{3}{2}+2\right)-\frac{4}{5}=\frac{13}{15}\)
\(x.\frac{7}{2}=\frac{13}{15}+\frac{4}{5}\)
\(x.\frac{5}{2}=\frac{13}{15}+\frac{12}{15}\)
\(x.\frac{7}{2}=\frac{25}{15}=\frac{5}{3}\)
\(x=\frac{5}{3}:\frac{7}{2}\)
\(x=\frac{5}{3}.\frac{2}{7}=\frac{10}{21}\)
a) =\(\left[\left(12+1\right)^2+\left(12+2\right)^2\right]:\left(13^2+14^2\right)\)
=1
b)=(1.2.3....8).(9-1-8)
=(1.2.3....8).0
=0
mik chỉ giải được zậy thôi.
t mik nha.
Mn giải có cách giải giúp mình nhé!
Bằng 2