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a) \(\left(\frac{-1}{6}+\frac{5}{-12}\right)+\frac{7}{12}=\left(\frac{-2}{12}+\frac{-5}{12}\right)+\frac{7}{12}=\left(\frac{-7}{12}\right)+\frac{7}{12}=0\)
b)\(\frac{7}{36}-\frac{8}{-9}+\frac{-2}{3}=\frac{7}{36}+\frac{32}{36}-\frac{24}{36}=\frac{15}{36}=\frac{5}{12}\)
c) \(\frac{3}{5}-\frac{2}{5}.\frac{10}{12}=\frac{3}{5}-\frac{2}{5}.\frac{5}{6}=\frac{3}{5}-\frac{1}{3}=\frac{9}{15}-\frac{5}{15}=\frac{4}{15}\)
d) \(\frac{2}{\left(-3\right)^2}+\frac{5}{-13}-\frac{-3}{4}=\frac{2}{9}-\frac{5}{13}+\frac{3}{4}=\frac{8}{36}-\frac{15}{36}+\frac{27}{36}=\frac{5}{9}\)
Lời giải:
$\frac{1}{5^2}+\frac{1}{6^2}+\frac{1}{7^2}+...+\frac{1}{100^2}$
$< \frac{1}{4.5}+\frac{1}{5.6}+\frac{1}{6.7}+...+\frac{1}{99.100}$
$=\frac{1}{4}-\frac{1}{5}+\frac{1}{5}-\frac{1}{6}+\frac{1}{6}-\frac{1}{7}+....+\frac{1}{99}-\frac{1}{100}$
$=\frac{1}{4}-\frac{1}{100}< \frac{1}{4}$
\(1,\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=\frac{12}{15}+\frac{12}{35}+\frac{12}{63}+\frac{12}{99}=6\left(\frac{2}{15}+\frac{2}{35}+\frac{2}{63}+\frac{2}{99}\right)=6\left(\frac{2}{3.5}+\frac{2}{5.7}+\frac{2}{7.9}+\frac{2}{9.11}\right).Tacocongthuc:\frac{1}{n}-\frac{1}{n+k}=\frac{k}{n\left(n+k\right)}\Rightarrow\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}=6\left(\frac{1}{3}-\frac{1}{5}+\frac{1}{5}-.....-\frac{1}{11}\right)=6\left(\frac{1}{3}-\frac{1}{11}\right)=\frac{48}{33}=\frac{16}{11}\)
\(2,\left(x+1\right)+\left(x+2\right)+.....+\left(x+211\right)=211x+\left(1+2+....+211\right)=211x+\frac{212.211}{2}=211x+22366=23632\Leftrightarrow211x=23632-22366=1266\Leftrightarrow x=6\)
a, \(14:\left(4\frac{2}{3}:1\frac{5}{9}\right)+14:\left(\frac{2}{3}+\frac{8}{9}\right)\)
=> \(14:\frac{28}{9}+14:\frac{14}{9}=>14.\frac{9}{28}+14.\frac{9}{14}\)
=> 14. ( \(\frac{9}{28}+\frac{9}{14}\) )
=> \(14.\frac{27}{28}=\frac{419}{28}\)
b, \(\frac{1212}{1515}+\frac{1212}{3535}+\frac{1212}{6363}+\frac{1212}{9999}\)
=> \(\frac{4}{5}+\frac{12}{35}+\frac{4}{21}+\frac{4}{33}\)
=> \(\frac{8}{7}+\frac{24}{77}=\frac{16}{11}\)
bài 2 :
( x + 1 ) + ( x + 2 ) + ... + ( x + 211 ) = 23632
=> ( x + x + x + ... + x ) + ( 1 + 2 + 3 + ... + 211 ) = 23632
=> 211x + 22366 = 23632
=> 211x = 23632 - 22366
=> 211x = 1266
=> x = 1266 : 211
x = 6
\(1)A=\frac{\frac{2}{5}+\frac{2}{7}-\frac{2}{9}-\frac{2}{11}}{\frac{4}{5}+\frac{4}{7}-\frac{4}{9}-\frac{4}{11}}\)
\(=\frac{2\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}{4\left(\frac{1}{5}+\frac{1}{7}-\frac{1}{9}-\frac{1}{11}\right)}\)
\(=\frac{2}{4}=\frac{1}{2}\)
\(2)B=\frac{1^2}{1.2}.\frac{2^2}{2.3}.\frac{3^2}{3.4}.\frac{4^2}{4.5}\)
\(=\frac{1.1}{1.2}.\frac{2.2}{2.3}.\frac{3.3}{3.4}.\frac{4.4}{4.5}\)
\(=\frac{1}{2}.\frac{2}{3}.\frac{3}{4}.\frac{4}{5}\)
\(=\frac{1.2.3.4}{2.3.4.5}=\frac{1}{5}\)
\(3)C=\frac{2^2}{1.3}.\frac{3^2}{2.4}.\frac{4^2}{3.5}.\frac{5^2}{4.6}\)
\(=\frac{2.2.3.3.4.4.5.5}{1.3.2.4.3.5.4.6}\)
\(=\frac{2.5}{1.6}=\frac{2.5}{1.3.2}=\frac{5}{3}\)
\(4)D=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{1}{5}-\frac{1}{6}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right)\left(\frac{6}{30}-\frac{5}{30}-\frac{1}{30}\right)\)
\(=\left(\frac{150}{1111}+\frac{5}{75}-\frac{14}{77}\right).0=0\)
\(5)M=8\frac{2}{7}-\left(3\frac{4}{9}+3\frac{9}{7}\right)\) \(N=\left(10\frac{2}{9}+2\frac{3}{5}\right)-6\frac{2}{9}\)
\(=\frac{58}{7}-\left(\frac{31}{9}+\frac{30}{7}\right)\) \(=\left(\frac{92}{9}+\frac{13}{5}\right)-\frac{56}{9}\)
\(=\frac{58}{7}-\left(\frac{217}{63}+\frac{270}{63}\right)\) \(=\left(\frac{460}{45}+\frac{117}{45}\right)-\frac{280}{45}\)
\(=\frac{58}{7}-\frac{487}{63}\) \(=\frac{577}{45}-\frac{280}{45}\)
\(=\frac{522}{63}-\frac{487}{63}=\frac{5}{9}\) \(=\frac{33}{5}\)
\(P=M-N\)
\(\Rightarrow P=\frac{5}{9}-\frac{33}{5}\)
\(\Rightarrow P=\frac{25}{45}-\frac{297}{45}\)
\(\Rightarrow P=\frac{-272}{45}\)
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 !