\(\frac{7}{8}\div\left(\frac{14}{3}+\frac{14}{4}\right)+\frac{10}{14}\)

\(=\frac{7}{8}\div\left(\frac{56}{12}+\frac{52}{12}\right)+\frac{10}{14}\)

\(=\frac{7}{8}\div\frac{108}{12}+\frac{10}{14}\)

\(=\frac{7}{8}\div9+\frac{5}{7}\)

\(=\frac{7}{8}\times\frac{1}{9}+\frac{5}{7}\)

\(=\frac{7}{72}+\frac{5}{7}\)

\(=\frac{49}{504}+\frac{360}{504}\)

\(=\frac{409}{504}\)

\(\frac{30}{42}\div\left(\frac{5}{7}-\frac{5}{21}\right)+\frac{6}{8}\)

\(=\frac{6}{7}\div\left(\frac{15}{21}-\frac{5}{21}\right)+\frac{3}{4}\)

\(=\frac{6}{7}\div\frac{10}{21}+\frac{3}{4}\)

\(=\frac{6}{7}\times\frac{21}{10}+\frac{3}{4}\)

\(=\frac{126}{70}+\frac{3}{4}\)

\(=\frac{504}{280}+\frac{210}{280}\)

\(=\frac{714}{280}\)

\(\left(6-\frac{14}{15}\right).\frac{50}{38}-\frac{8}{18}\)

\(=\left(\frac{90}{15}-\frac{14}{15}\right).\frac{25}{19}-\frac{4}{9}\)

\(=\frac{76}{15}.\frac{25}{19}-\frac{4}{9}\)

\(=\frac{76.25}{15.19}-\frac{4}{9}\)

\(=\frac{19.2.2.5.5}{3.5.19}-\frac{4}{9}\)

\(=\frac{20}{3}-\frac{4}{9}\)

\(=\frac{60}{9}-\frac{4}{9}\)

\(=\frac{56}{9}\)

\(\frac{4}{6}+\frac{5}{21}.\left(\frac{4}{5}-\frac{2}{6}\right)\)

\(=\frac{2}{3}+\frac{5}{21}.\left(\frac{24}{30}-\frac{10}{30}\right)\)

\(=\frac{2}{3}+\frac{5}{21}.\frac{14}{30}\)

\(=\frac{2}{3}+\frac{70}{630}\)

\(=\frac{420}{630}+\frac{70}{630}\)

\(=\frac{490}{630}\)

\(=\frac{7}{9}\)

\(\Leftrightarrow\dfrac{1}{2};\dfrac{1}{2};\dfrac{8}{5};\dfrac{3}{2}\)

14/28=1/2

25/50=1/2

48/30=8/5

81/54=3/2

\(\dfrac{14}{28}=\dfrac{1}{2}\)

\(\dfrac{25}{50}=\dfrac{1}{2}\)

\(\dfrac{48}{30}=\dfrac{8}{5}\)

\(\dfrac{81}{54}=\dfrac{3}{2}\)

B=2^10*(2*3)^15+3^14*3*5*(2^2)^13/2^19*(2*3^2)^7*3-3^15*2^25

B=2^10*2^15*3^15+3^15+3^15*5*2^26/2^19*2^7*3^14**3-3^15*2^25

B=2^25*3^15+5*2*3^15*2^25/2*2^25*3^15-3^15*2^25

B=1*2^25*3^15+10*2^25*3^15/2*2^25*3^15-2^25*3^15

B=2^25*3^15*(1+10)/2^25*3^15*(2-1)=11/1=11

cái hàng cuối có dấu chấm đằng trước bỏ nhé cảm ơn

a)12/3
b)8/3
c)43/19
d)5/11

12/3 , 8/3 ,43/19 , 5/11

Ta có : \(14-\left(25-x\right)+\left(-14+50\right).14-\left(25-x\right)+\left(-14+50\right)\)

\(=14-25+x+36.14-25+x-14+50\)

\(=2x+36.14=2x+504\)

Giải:

\(14-\left(25-x\right)+\left(-14+50\right).14-\left(25-x\right)+\left(-14+50\right)\) 

\(=14-25+x+\left(-14.14+50.14\right)-25+x-14+50\) 

\(=\left(14-14\right)+\left(-25-25+50\right)+\left(x+x\right)+\left(-196+700\right)\) 

\(=0+0+2x+504\) 

\(=2x+504\) 

Chúc bạn học tốt!

a) \(\dfrac{7}{56}=\dfrac{7:7}{56:7}=\dfrac{1}{8}\)

b) \(\dfrac{25}{100}=\dfrac{25:25}{100:25}=\dfrac{1}{4}\)

a) 7/56=7:7/56:7=1/8

b) 25/100=25:25/100:25=1/4

25100=25:25100:25=14

ta có: \(E=\frac{50^{12}.16^{14}}{8^{25}.125^8}\)

\(E=\frac{\left(2.5^2\right)^{12}.\left(2^4\right)^{14}}{\left(2^3\right)^{25}.\left(5^3\right)^8}\)

\(E=\frac{2^{68}.5^{24}}{2^{75}.5^{24}}=\frac{1}{2^7}=\frac{1}{128}\)

ta có: \(E=\frac{50^{12}.16^{14}}{8^{25}.125^8}\)

\(E=\frac{\left(2.5^2\right)^{12}.\left(2^4\right)^{14}}{\left(2^3\right)^{25}.\left(5^3\right)^8}\)

\(E=\frac{2^{68}.5^{24}}{2^{75}.5^{24}}=\frac{1}{2^7}=\frac{1}{128}\)

Ta có : \(E=\frac{50^{12}.16^{14}}{8^{25}.125^8}=\frac{\left(2.5^2\right)^{12}.\left(2^4\right)^{14}}{\left(2^3\right)^{25}.\left(5^3\right)^8}\)

                \(=\frac{2^{12}.5^{24}.2^{56}}{2^{75}.5^{24}}=\frac{2^{68}.5^{24}}{2^{75}.5^{24}}\)

                 \(=\frac{1}{2^7}=\frac{1}{128}\)

Chúc bạn học tốt !

14/28=1/2

25/50 = 1/2

48/30= 8/5

20/30 = 2/3

14/28=1/2

15/20=1/2

48/30=8/5

20/30=2/3

Ta có:\(\frac{14}{28}=\frac{14:14}{28:14}=\frac{1}{2}\)

\(\frac{25}{50}=\frac{25:25}{50:25}=\frac{1}{2}\)

\(\frac{48}{30}=\frac{48:6}{30:6}=\frac{8}{5}\)

\(\frac{20}{30}=\frac{20:10}{30:10}=\frac{2}{3}\)

\(\frac{25^{14}.5^{10}.625^3.125^7}{5^{14}.125^{10}.25^3.625^7}=\frac{\left(5^2\right)^{14}.5^{10}.\left(5^4\right)^3.\left(5^3\right)^7}{5^{14}.\left(5^3\right)^{10}.\left(5^2\right)^3.\left(5^4\right)^7}=\frac{5^{28}.5^{10}.5^{12}.5^{21}}{5^{14}.5^{30}.5^6.5^{28}}\)

\(=\frac{5^{71}}{5^{78}}=\frac{1}{5^7}=\frac{1}{78125}\)