biểu thức đâu

a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)

b:

Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)

 \(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)

\(=71-43+1+1\)

=28+2=30

khó quá

a: \(A=49+\dfrac{8}{23}-14-\dfrac{8}{23}-5-\dfrac{7}{32}=30-\dfrac{7}{32}=\dfrac{953}{32}\)

b:

Sửa đề: \(B=71\dfrac{38}{45}-\left(43\dfrac{8}{45}-1\dfrac{17}{51}\right)\)

 \(B=71+\dfrac{38}{45}-43-\dfrac{8}{45}+1+\dfrac{17}{51}\)

\(=71-43+1+1\)

=28+2=30

a: \(A=\dfrac{25^6}{5^3}=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)

b: \(B=32\cdot\left(\dfrac{3}{2}\right)^5=32\cdot\dfrac{3^5}{2^5}=32\cdot\dfrac{243}{32}=243\)

c: \(C=\left(\dfrac{1}{3}\right)^4\cdot3^{-3}=3^{-4}\cdot3^{-3}=3^{-4-3}=3^{-7}\)

d: \(D=4^{-2}\cdot\left(\dfrac{2}{5}\right)^5\cdot5^4\)

\(=\dfrac{1}{4^2}\cdot\dfrac{2^5}{5^5}\cdot5^4\)

\(=\dfrac{1}{16}\cdot\dfrac{32}{5}=\dfrac{2}{5}\)

e: \(E=9^{-5}:\left(\dfrac{5}{3}\right)^4\cdot25^2\)

\(=\dfrac{1}{9^5}:\dfrac{5^4}{3^4}\cdot\left(5^2\right)^2\)

\(=\dfrac{1}{3^{10}}\cdot\dfrac{3^4}{5^4}\cdot5^4=\dfrac{1}{3^6}\)

f: \(F=\left(\dfrac{5}{8}\right)^{-2}:4^2\)

\(=\left(1:\dfrac{5}{8}\right)^2:4^2\)

\(=\left(\dfrac{8}{5}\right)^2\cdot\dfrac{1}{16}=\dfrac{64}{25}\cdot\dfrac{1}{16}=\dfrac{4}{25}\)

g: \(G=\left(\dfrac{5}{3}\right)^3\cdot\left(\dfrac{9}{2}\right)^2:\left(\sqrt{3}\right)^4\)

\(=\dfrac{5^3}{3^3}\cdot\dfrac{9^2}{2^2}:9\)

\(=\dfrac{5^3\cdot3^4}{3^3\cdot2^2}\cdot\dfrac{1}{3^2}\)

\(=\dfrac{125}{2^2\cdot3}=\dfrac{125}{3\cdot4}=\dfrac{125}{12}\)

\(A=\dfrac{\left(5^2\right)^6}{5^3}=\dfrac{5^{12}}{5^3}=5^9\)

\(B=32.\left(\dfrac{3}{2}\right)^5=\dfrac{2^5.3^5}{2^5}=2^5\)

\(C=\left(\dfrac{1}{3}\right)^4.3^{-3}=\dfrac{1}{3^4.3^3}=\dfrac{1}{3^7}\)

\(D=4^{-2}.\left(\dfrac{2}{5}\right)^5.5^4=\dfrac{1}{\left(2^2\right)^2}.\dfrac{2^5}{5^5}.5^4=\dfrac{2}{5}\)

\(E=\dfrac{1}{9^5}.\dfrac{3^4}{5^4}.\left(5^2\right)^2=\dfrac{1}{3^{10}}.\dfrac{3^4}{5^4}.5^4=\dfrac{1}{3^6}\)

\(F=\dfrac{8^2}{5^2}:\left(2^2\right)^2=\dfrac{\left(2^3\right)^2}{5^2.2^4}=\dfrac{2^6}{5^2.2^4}=\dfrac{2^2}{5^2}\)

\(G=\dfrac{5^3}{3^3}.\dfrac{\left(3^2\right)^2}{2^2}:3^2=\dfrac{5^3}{3^3}.\dfrac{3^4}{2^2}.\dfrac{1}{3^2}=\dfrac{5^3}{3.2^2}\)

\(M=\left(\dfrac{1}{3}\right)^{12}\cdot\left(\dfrac{1}{3}\right)^{-15}+\left(\dfrac{2}{5}\right)^{-4}\cdot5^{-4}\cdot32\)

\(=\left(\dfrac{1}{3}\right)^{-3}+2^{-4}\cdot32\)

\(=27+\dfrac{32}{16}=27+2=29\)

  \(=49\frac{8}{23}-5\frac{7}{32}-14\frac{8}{23}\)

\(=\left(49\frac{8}{23}-14\frac{8}{23}\right)-5\frac{7}{32}\)

\(=35\frac{0}{23}-5\frac{7}{32}\)

\(=\frac{805}{23}-\frac{167}{32}\)

\(\frac{25760}{736}-\frac{3841}{736}\)

\(=\frac{21919}{736}=\frac{953}{32}\)

1135/23-167/32+330/23

=(1135/23-330/23)+167/32

=805/23+167/32

=35+167/32

=1120/32+167/32

=1287/32

\(\begin{array}{l}B = \left( {\frac{{ - 3}}{{13}}} \right) + \frac{{16}}{{23}} + \left( {\frac{{ - 10}}{{13}}} \right) + \frac{5}{{11}} + \frac{7}{{23}}\\ = \left[ {\left( {\frac{{ - 3}}{{13}}} \right) + \left( {\frac{{ - 10}}{{13}}} \right)} \right] + \left[ {\frac{{16}}{{23}} + \frac{7}{{23}}} \right] + \frac{5}{{11}}\\ =  - 1 + 1 + \frac{5}{{11}}\\ = \frac{5}{{11}}\end{array}\)

`B= ( (-3)/13 + (-10)/13) + (16/23 + 7/23 ) +5/11`

`B= -13/13 + 23/23 +5/11`

`B=-1+1+5/11`

`B=0+5/11`

`B=5/11`

\(B=\left(-\dfrac{3}{13}\right)+\dfrac{16}{23}+\left(-\dfrac{10}{13}\right)+\dfrac{5}{11}+\dfrac{7}{23}\)

\(B=\left[\left(-\dfrac{3}{13}\right)+\left(-\dfrac{10}{13}\right)\right]+\left(\dfrac{16}{23}+\dfrac{7}{23}\right)+\dfrac{5}{11}\)

\(B=\left(-1\right)+1+\dfrac{5}{11}\)

\(B=\dfrac{5}{11}\)

\(A=\left(4+\frac{1}{5}\right).\frac{19}{8}+\left(2+\frac{5}{8}\right).\frac{21}{5}\)                                                                                                                                                     =\(\frac{21}{5}.\frac{19}{8}+\frac{21}{8}.\frac{21}{5}\)                                                                                                                                                                             =\(\frac{21}{5}.\left(\frac{19}{8}+\frac{21}{8}\right)\)                                                                                                                                                                           = \(\frac{21}{5}.5\) =21                                                                                                                                                                                       \(B=\frac{25}{2}.\left(3+\frac{2}{7}\right)-\frac{23}{7}.\left(5+\frac{1}{2}\right)\)                                                                                                                                            =\(\frac{25}{2}.\frac{23}{7}-\frac{23}{7}.\frac{11}{2}\)                                                                                                                                                                       =\(\frac{23}{7}.\left(\frac{25}{2}-\frac{11}{2}\right)\)                                                                                                                                                                          =\(\frac{23}{7}.7=23\)

\(a,\left(7+3\dfrac{1}{4}-\dfrac{3}{5}\right)+\left(0,4-5\right)-\left(4\dfrac{1}{4}-1\right)\)

\(=\left(7+\dfrac{13}{4}-\dfrac{3}{5}\right)-\dfrac{23}{5}-\left(\dfrac{17}{4}-1\right)\)

\(=7+\dfrac{13}{4}-\dfrac{3}{5}-\dfrac{23}{5}-\dfrac{17}{4}+1\)

\(=\left(7+1\right)+\left(\dfrac{13}{4}-\dfrac{17}{4}\right)-\left(\dfrac{3}{5}+\dfrac{23}{5}\right)\)

\(=8-\dfrac{4}{4}-\dfrac{26}{5}\)

\(=7-\dfrac{26}{5}\)

\(=\dfrac{9}{5}\)

\(b,\dfrac{2}{3}-\left[\left(-\dfrac{7}{4}\right)-\left(\dfrac{1}{2}+\dfrac{3}{8}\right)\right]\)

\(=\dfrac{2}{3}-\left(-\dfrac{7}{4}-\dfrac{1}{2}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{14}{8}-\dfrac{4}{8}-\dfrac{3}{8}\right)\)

\(=\dfrac{2}{3}-\left(-\dfrac{21}{8}\right)\)

\(=\dfrac{2}{3}+\dfrac{21}{8}\)

\(=\dfrac{79}{24}\)

\(c,\left(9-\dfrac{1}{2}-\dfrac{3}{4}\right):\left(7-\dfrac{1}{4}-\dfrac{5}{8}\right)\)

\(=\left(\dfrac{36}{4}-\dfrac{2}{4}-\dfrac{3}{4}\right):\left(\dfrac{56}{8}-\dfrac{2}{8}-\dfrac{5}{8}\right)\)

\(=\dfrac{31}{4}:\dfrac{49}{8}\)

\(=\dfrac{62}{49}\)

\(d,3-\dfrac{1-\dfrac{1}{7}}{1+\dfrac{1}{7}}=3-\dfrac{\dfrac{7}{7}-\dfrac{1}{7}}{\dfrac{7}{7}+\dfrac{1}{7}}=3-\left(\dfrac{6}{7}:\dfrac{8}{7}\right)=3-\dfrac{3}{4}=\dfrac{9}{4}\)