\(\text{​​}\text{​​}\)\(=\frac{27}{43}.\frac{34}{58}-\frac{21}{41}.\frac{1}{2}+\frac{9}{58}:\frac{43}{37}-\frac{6}{29}:\frac{41}{21}\\ =\frac{27}{43}.\frac{34}{58}-\frac{21}{41}.\frac{1}{2}+\frac{9}{58}.\frac{37}{43}-\frac{6}{29}.\frac{21}{41}\)

\(\frac{6}{43}\)

a) -90/189 + 45/84 - 78/126

= -10/21 + 15/28 - 13/21

= (-10/21 - 13/21) + 15/28

= -24/21 + 15/28

= -17/28

\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}=(\frac{-5}{12}+\frac{17}{12})+(\frac{4}{37}-\frac{41}{37})=\frac{12}{12}+\frac{-37}{37}=1+(-1)=0\)

\(B=\frac{1}{2}-\frac{43}{101}+\frac{-1}{3}-\frac{1}{6}=\frac{-43}{101}+(\frac{1}{2}+\frac{-1}{3}-\frac{1}{6})=\frac{-43}{101}+(\frac{3}{6}+\frac{-2}{6}-\frac{1}{6})=\frac{-43}{101}+0=\frac{-43}{101}\)

\(A=\frac{-5}{12}+\frac{4}{37}+\frac{17}{12}-\frac{41}{37}.\)

\(A=\left(\frac{-5}{12}+\frac{17}{12}\right)-\left(\frac{41}{37}-\frac{4}{37}\right)\)

\(A=1-1=0\)

\(B=\frac{1}{2}-\frac{43}{101}+\left(\frac{-1}{3}\right)-\frac{1}{6}\)

\(B=\left(\frac{1}{2}+\left(\frac{-1}{3}\right)-\frac{1}{6}\right)-\frac{43}{101}\)

\(A=0-\frac{43}{101}=\frac{-43}{101}\)

\(C=\frac{-5}{6}\cdot\frac{12}{-7}\cdot-\frac{21}{15}\)

\(C=\frac{-5}{2.3}\cdot\frac{3.2.2}{-7}\cdot\frac{3.\left(-7\right)}{3.5}\)

\(C=\frac{-2}{1}=-2\)

a. \(25^3:5^2\)
\(=\left(5^2\right)^3:5^2\)
\(=5^6:5^2=5^4\)
b. \(\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)
\(=\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)
\(=\left(\frac{3}{7}\right)^{21-\left(2+6\right)}=\left(\frac{3}{7}\right)^{21-12}=\left(\frac{3}{7}\right)^9\)

\(a,25^3:5^2\)

=\(\left(5^2\right)^3:5^2\)

=\(5^6:5^2\)

=\(5^4\)

\(b,\left(\frac{3}{7}\right)^{21}:\left(\frac{9}{49}\right)^6\)

=\(\left(\frac{3}{7}\right)^{21}:\left[\left(\frac{3}{7}\right)^2\right]^6\)

\(=\left(\frac{3}{7}\right)^{21}:\left(\frac{3}{7}\right)^{12}\)

\(=\left(\frac{3}{7}\right)^9\)

\(c,3-\left(\frac{6}{7}\right)^0+\left(\frac{1}{2}\right)^2:2\)

=\(3-1+\frac{1}{4}:2\)

\(=2+\frac{1}{4}\cdot\frac{1}{2}\)

\(=2+\frac{1}{8}\)

\(=\frac{17}{8}\)

\(d,\left(-\frac{7}{4}:\frac{5}{8}\right)\cdot\frac{11}{16}\)

\(=\left(-\frac{7}{4}\cdot\frac{8}{5}\right)\cdot\frac{11}{16}\)

\(=-\frac{14}{5}\cdot\frac{11}{16}\)

\(=-\frac{77}{40}\)

\(e,\frac{2}{3}+\frac{1}{3}\cdot\frac{-6}{10}\)

\(=\frac{2}{3}-\frac{1}{5}\)

\(=\frac{7}{15}\)

a/ Ta có:

\(\frac{3}{124}=\frac{30}{1240}\) ; \(\frac{1}{41}=\frac{30}{1230}\) ; \(\frac{5}{207}=\frac{30}{1242}\) ; \(\frac{2}{83}=\frac{30}{1245}\)

Vì các phân số trên đều cùng tử nên ta so sánh mẫu : 1230<1240<1242<1242

                                                  => \(\frac{30}{1230}>\frac{30}{1240}>\frac{30}{1242}>\frac{30}{1245}\)

                                                  Hay : \(\frac{1}{41}>\frac{3}{124}>\frac{5}{207}>\frac{2}{83}\)

b/ Ta có: 

   \(\frac{16}{9}=\frac{48}{27};\frac{24}{13}=\frac{48}{26}\)

Vì 27>26

=> \(\frac{16}{9}< \frac{24}{13}\)

; ; ;

Vì các phân số trên đều cùng tử nên ta so sánh mẫu : 1230<1240<1242<1242

=>

Hay :

b/ Ta có:

Vì 27>26

=>

              Bài làm :

a)\(=-\frac{3}{5}+\frac{28}{5}\times\frac{9}{14}=-\frac{3}{5}+\frac{18}{5}=3\)

b)\(=\frac{55}{126}+\frac{5}{42}+\frac{4}{9}=1\)

c)\(=-\frac{51}{13}-\frac{27}{13}=-6\)

d)\(=\frac{7}{3}-11\frac{1}{4}\times\frac{2}{15}=\frac{7}{3}-\frac{3}{2}=\frac{5}{6}\)

e)\(=1\times\frac{8}{3}\times0,25=\frac{2}{3}\)