Задание № 1025
Вычислите, используя распределительный закон:
а) $1\frac13\ast2=(1+\frac13)\ast2=1\ast2+\frac13\ast2=2+\frac23=2\frac23$
б) $1\frac15\ast2=(1+\frac15)\ast2=1\ast2+\frac15\ast2=2+\frac25=2\frac25$
в) $2\frac15\ast3=(2+\frac15)\ast3=2\ast3+\frac15\ast3=6+\frac35=6\frac35$
г) $3\frac14\ast3=(3+\frac14)\ast3=3\ast3+\frac14\ast3=9+\frac34=9\frac34$
д) $2\frac27\ast3=(2+\frac27)\ast3=2\ast3+\frac27\ast3=6+\frac67=6\frac67$
е) $2\ast5\frac14=2\ast(5+\frac14)=2\ast5+2\ast\frac14=10+\frac24=10\frac24=10\frac12$
ж) $2\frac49\ast9=(2+\frac49)\ast9=2\ast9+\frac49\ast9=18+4=22$
з) $2\ast5\frac78=2\ast(5+\frac78)=2\ast5+2\ast\frac78=10+\frac74=10+1\frac34=11\frac34$
и) $2\frac19\ast3=(2+\frac19)\ast3=2\ast3+\frac19\ast3=6+\frac13=6\frac13$
Задание № 1026
Вычислите, используя распределительный закон:
а) $\frac13\ast2+2\frac23\ast2=(\frac13+2\frac23)\ast2=3\ast2=6$
б) $1\frac15\ast3+\frac45\ast3=(1\frac15+\frac45)\ast3=1\frac55\ast3=2\ast3=6$
в) $2\frac45\ast3-\frac45\ast3=(2\frac45-\frac45)\ast3=2\ast3=6$
г) $2\frac45\ast\frac59+6\frac15\ast\frac59=(2\frac45+6\frac15)\ast\frac59=8\frac55\ast\frac59=9\ast\frac59=5$
д) $2\frac45\ast\frac58-1\frac15\ast\frac58=(2\frac45-1\frac15)\ast\frac58=1\frac35\ast\frac58=\frac85\ast\frac58=1$
е) $3\frac15\ast\frac8{11}+3\frac15\ast\frac3{11}=3\frac15\ast(\frac8{11}+\frac3{11})=3\frac15\ast\frac{11}{11}=3\frac15$
ж) $4\frac23\ast7\frac1{12}-6\frac1{12}\ast4\frac23=(7\frac1{12}-6\frac1{12})\ast4\frac23=1\ast4\frac23=4\frac23$
Задание № 1027
Вычислите:
а) $4\frac1{12}\ast8\frac67\ast6=\frac{7\ast7\ast2\ast31\ast6}{2\ast6\ast7}=\frac{7\ast31}1=217$
б) $6\frac14\ast1\frac25\ast8=\frac{25}4\ast\frac75\ast8=\frac{5\ast5\ast7\ast4\ast2}{4\ast5}=\frac{5\ast7\ast2}1=70$
в) $5\frac37\ast\frac59\ast\frac{14}{19}=\frac{38}7\ast\frac59\ast\frac{14}{19}=\frac{19\ast2\ast5\ast7\ast2}{7\ast9\ast19}=\frac{2\ast5\ast2}9=\frac{20}9=2\frac29$
г) $3\frac38\ast\frac12:\frac9{16}=\frac{27}8\ast\frac12\ast\frac{16}9=\frac{9\ast3\ast8\ast2}{8\ast2\ast9}=\frac31=3$
д) $5\frac57\ast\frac5{12}\ast5\frac14\ast\frac67:1\frac12=\frac{40}7\ast\frac5{12}\ast\frac{21}4\ast\frac67\ast\frac23=\frac{5\ast4\ast2\ast5\ast21\ast6\ast2}{7\ast4\ast3\ast4\ast21}=\frac{5\ast2\ast5}7=\frac{50}7=7\frac17$
е) $\frac34\ast1\frac14\ast7\frac12\ast2\frac14:7\frac12=\frac34\ast\frac54\ast\frac{15}2\ast\frac94\ast\frac{15}2=\frac{3\ast5\ast15\ast9\ast2}{4\ast4\ast2\ast4\ast15}=\frac{135}{64}=2\frac7{64}$
ж) $\frac37:1\frac17:1\frac78:\frac{11}{25}:1\frac1{22}:3\frac2{11}=\frac37\ast\frac78\ast\frac8{15}\ast\frac{25}{11}\ast\frac{22}{23}\ast\frac{11}{35}=\frac{3\ast7\ast8\ast5\ast5\ast22\ast11}{7\ast8\ast3\ast5\ast11\ast23\ast7\ast5}=\frac{22}{7\ast23}=\frac{22}{161}$
з) $\frac{15}{16}:1\frac1{48}:2\frac17:2\frac2{49}:\frac{24}{125}=\frac{15}{16}\ast\frac{48}{49}\ast\frac7{15}\ast\frac{49}{100}\ast\frac{125}{24}=\frac{15\ast2\ast24\ast7\ast49\ast5\ast25}{16\ast49\ast15\ast25\ast4\ast24}=\frac{7\ast5}{8\ast4}=\frac{35}{32}=1\frac3{32}$
Задание № 1028
Вычислите:
а) $(\frac2{15}+1\frac7{12})\ast\frac{30}{103}-2:2\frac14\ast\frac9{32}+2\frac13=2:2\frac14\ast\frac9{32}+2\frac13=(\frac8{60}+1\frac{35}{60})\ast\frac{30}{103}-\frac21\ast\frac49\ast\frac9{32}+2\frac13=1\frac{43}{60}\ast\frac{30}{103}-\frac89\ast\frac9{8\ast4}+2\frac13=\frac{103}{60}\ast\frac{30}{103}-\frac14+2\frac13=\frac12-\frac14+2\frac13=\frac12-\frac14+2\frac13=2+\frac{6-3+4}{12}=2\frac7{12}$
б) $(7\frac12\ast2\frac23-12\frac14:\frac79):6+3\frac18+5\frac27=(20-\frac{63}4)\ast\frac16+3\frac18+5\frac27=\frac{20}6-\frac{63}{4\ast6}+3\frac18+5\frac27=\frac{80}{24}-\frac{63}{24}+3\frac18+5\frac27=\frac{17}{24}+3\frac18+5\frac27=8+\frac{119+21+48}{168}=8+\frac{188}{168}=8+1\frac{20}{168}=9\frac5{42}$
в) $5\frac13:6\frac25+(12:3\frac35-\frac23)\ast\frac23+7\frac25=\frac{16}3\ast\frac5{32}+(\frac{12}1\ast\frac5{18}-\frac23)\ast\frac23+7\frac25=\frac56+(\frac{10}3-\frac23)\ast\frac23+7\frac25=\frac56+\frac83\ast\frac23+7\frac25=\frac56+\frac{16}9+7\frac25=\frac{15}{18}+\frac{32}{18}+7\frac25=\frac{47}{18}+7\frac25=7+\frac{235+36}{90}=7\frac{271}{90}=10\frac1{90}$
г) $21\frac2{59}-\frac25\ast(3\frac{15}{28}:\frac9{28}-1:1\frac{10}{49}):2=21\frac2{59}-\frac25\ast(\frac{99}{28}\ast\frac{28}9-1\ast\frac{49}{59})\ast\frac12=21\frac5{59}-\frac25\ast(11-\frac{49}{59})\ast\frac12=21\frac2{59}-\frac25\ast(10\frac{59}{59}-\frac{49}{59})\ast\frac12=21\frac2{59}-\frac25\ast10\frac{10}{59}\ast\frac12=21\frac2{59}-\frac25\ast\frac{600}{59}\ast\frac12=21\frac2{59}-\frac15\ast\frac{5\ast120}{59}=21\frac2{59}-\frac{120}{59}=19\frac{120}{59}-\frac{120}{59}=19$
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