\(\frac{4}{5}\):\(\frac{2}{3}\)-\(\frac{4}{5}\):\(\frac{6}{7}\)+\(\frac{4}{5}\):\(\frac{3}{14}\)
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Mình làm như thế này nek
\(\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{0,75+\frac{9}{7}-2\frac{2}{5}}+\frac{\frac{3}{14}-\frac{2}{10}+\frac{5}{18}+\frac{7}{66}}{\frac{6}{7}-\frac{4}{5}+\frac{10}{9}+\frac{14}{33}}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{\frac{2}{4}+\frac{9}{7}-\frac{12}{5}}+\frac{\frac{1}{2}\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}{2\cdot\left(\frac{3}{7}-\frac{2}{5}+\frac{5}{9}+\frac{7}{33}\right)}\)
\(=\frac{\frac{1}{4}+\frac{3}{7}-\frac{4}{5}}{3\cdot\left(\frac{1}{4}+\frac{3}{7}-\frac{4}{5}\right)}+\frac{\frac{1}{2}}{2}\)
\(=\frac{1}{3}+\frac{1}{4}=\frac{7}{12}\)

Bài 1:
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + (\(\frac45\) - \(\frac{3}{17}\) + \(\frac13\)) - \(\frac17\) + (- \(\frac{14}{30}\))
A = \(\frac15\) + \(\frac{3}{17}\) - \(\frac43\) + \(\frac45\) - \(\frac{3}{17}\) + \(\frac13\) - \(\frac17\) - \(\frac{14}{30}\)
A = (\(\frac15\) + \(\frac45\)) + (\(\frac{3}{17}\) - \(\frac{3}{17}\)) - (\(\frac43-\frac13\)) - \(\frac{30}{210}\) - \(\frac{98}{210}\)
A = 1 + 0 - 1 - (\(\frac{30}{210}+\frac{98}{210}\))
A = 1 - 1 - \(\frac{228}{210}\)
A = 0 - \(\frac{128}{210}\)
A = - \(\frac{64}{105}\)
Bài 2:
B= (\(\frac58\) - \(\frac{4}{12}\) + \(\frac32\)) - (\(\frac58\) + \(\frac{9}{13}\)) - (\(\frac{-3}{2}\)) + \(\frac{7}{-15}\)
B = \(\frac58\) - \(\frac{4}{12}\) + \(\frac32\) - \(\frac58\) - \(\frac{9}{13}\) + \(\frac32\) - \(\frac{7}{15}\)
B = (\(\frac58\) - \(\frac58\)) + (\(\frac32\) + \(\frac32\)) - (\(\frac13\) + \(\frac{9}{13}\) + \(\frac{7}{15}\))
B = 0 + 3 - (\(\frac{65}{195}\) + \(\frac{135}{195}\) + \(\frac{91}{195}\))
B = 3 - (\(\frac{200}{195}\) + \(\frac{91}{195}\))
B = 3 - \(\frac{97}{65}\)
B = \(\frac{195}{65}\) - \(\frac{97}{65}\)
B = \(\frac{98}{65}\)

a) \(\frac{14}{21}+1-\left|\frac{1}{3}-1\right|\)
\(=\frac{2}{3}+1-\frac{2}{3}\)
\(=1+\left(\frac{2}{3}-\frac{1}{3}\right)\)
\(=1\)
b) \(\frac{1}{3}-\left|\frac{-1}{4}+\frac{5}{6}\right|-\left|\frac{-7}{12}\right|\)
\(=\frac{1}{3}-\frac{7}{12}-\frac{7}{12}\)
\(=-\frac{5}{6}\)

\(a)\) \(A=\frac{5\left(2^2.3^2\right)^9.\left(2^2\right)^6-2\left(2^2.3\right)^{14}.3^4}{5.2^{28}.3^{18}-7.2^{29}.3^{18}}\)
\(A=\frac{2^{30}.3^{18}.5-2^{29}.3^{18}}{2^{28}.3^{18}.5-2^{29}.3^{18}.7}\)
\(A=\frac{2^{29}.3^{18}\left(2.5-1\right)}{2^{28}.3^{18}\left(5-2.7\right)}\)
\(A=\frac{2\left(10-1\right)}{5-14}\)
\(A=\frac{2.9}{-9}\)
\(A=-2\)
Vậy \(A=-2\)
\(b)\) \(B=81.\left[\frac{12-\frac{12}{7}-\frac{12}{289}-\frac{12}{85}}{4-\frac{4}{7}-\frac{4}{289}-\frac{4}{85}}:\frac{5+\frac{5}{13}+\frac{5}{169}+\frac{5}{91}}{6+\frac{6}{13}+\frac{6}{169}+\frac{6}{91}}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}{4\left(1-\frac{1}{7}-\frac{1}{289}-\frac{1}{85}\right)}:\frac{5\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}{6\left(1+\frac{1}{13}+\frac{1}{169}+\frac{1}{91}\right)}\right].\frac{158158158}{711711711}\)
\(B=81.\left[\frac{12}{4}:\frac{5}{6}\right].\frac{2}{9}\)
\(B=81.\frac{18}{5}.\frac{2}{9}\)
\(B=\frac{324}{5}\)
Vậy \(B=\frac{324}{5}\)
Chúc bạn học tốt ~ ( mỏi tay qué >_< )
\(\frac{4}{5}:\frac{2}{3}-\frac{4}{5}:\frac{6}{7}+\frac{4}{5}:\frac{3}{14}\)=\(\frac{4}{5}\cdot\frac{3}{2}-\frac{4}{5}\cdot\frac{7}{6}+\frac{4}{5}\cdot\frac{14}{3}\)=\(\frac{4}{5}\cdot\left(\frac{3}{2}-\frac{7}{6}+\frac{14}{3}\right)\)=\(\frac{4}{5}\cdot5\)=\(4\)
bằng 4 nhé