Алгебра, вопрос задал sobollarsenii , 1 год назад

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Ответил NNNLLL54

Ответ:

$\bf \displaystyle 11)\ \ \Big(\frac{3}{4}+\frac{1}{3}:2\Big):10=\Big[2+\Big(1-\frac{1}{2}\Big)^3:\Big(1-\frac{3}{8}\Big)\Big]:x$

Упростим выражения в правой и левой частях равенства .

$\bf \displaystyle \Big(\frac{3}{4}+\frac{1}{6}\Big):10=\Big[2+\Big(\frac{1}{2}\Big)^3:\Big(\frac{5}{8}\Big)\Big]:x\\\\\\\frac{11}{12}:10=\Big[2+\frac{1}{8}\cdot \frac{8}{5}\Big]:x\\\\\\\frac{11}{120}=\Big[2+\frac{1}{5}\Big]:x\ \ \ ,\ \ \ \ \frac{11}{120}=\frac{11}{5}:x\ \ \ ,\ \ \ x=\frac{11}{5}:\frac{11}{120}\ \ ,\\\\\\x=\frac{11}{5}\cdot \frac{120}{11}\ \ ,\ \ \ x=\frac{120}{5}\ \ \ ,\ \ \ x=24\\\\\\Otvet:\ x=24\ .$

$\bf \displaystyle 12)\ \ \Big(\frac{1}{3}+\frac{7}{14}\cdot \frac{8}{5}-\frac{1}{15}\Big): \Big(1+\frac{1}{3}\Big)=\Big(5+\frac{1}{3}\Big):x\\\\\\\Big(\frac{1}{3}+\frac{4}{5}-\frac{1}{15}\Big): \Big(\frac{4}{3}\Big)=\frac{16}{3}:x\\\\\\\frac{5+12-1}{15}\cdot\frac{3}{4}=\frac{16}{3}:x\\\\\\\frac{16}{15}\cdot \frac{3}{4}=\frac{16}{3}:x\ \ \ ,\ \ \ \frac{4}{5}=\frac{16}{3}:x\ \ \ ,\ \ \ x=\frac{16}{3}:\frac{4}{5}\ \ \ ,\ \ \ x=\frac{16}{3}\cdot \frac{5}{4}\ \ ,$

$\bf x=\dfrac{4\cdot 5}{3}\ \ \ ,\ \ \ x=\dfrac{20}{3}\ \ \ ,\ \ \ x=6\dfrac{2}{3}\\\\\\Otvet:\ \ x=6\dfrac{2}{3} \ .$

$\bf \displaystyle 13)\ \ x:\Big[\ 2+\frac{1}{2}-\Big(1-\frac{1}{2}\Big)^2\Big]=\Big(1-\frac{1}{8}\Big):\Big(1+\frac{1}{2}-\frac{2}{7}\cdot \frac{21}{16}\Big)\\\\\\x:\Big[\ \frac{5}{2}-\frac{1}{4}\ \Big]=\frac{7}{8}:\Big(\frac{3}{2}-\frac{3}{8}\Big)\\\\\\x:\frac{9}{4}\ =\frac{7}{8}:\frac{9}{8}\ \ \ ,\ \ \ x:\frac{9}{4}=\frac{7}{9}\ \ \ ,\ \ \ x=\frac{7}{9}\cdot \frac{9}{4}\ \ \ ,\ \ \ x=\frac{7}{4}\ \ ,\ \ \ x=1\frac{3}{4}\\\\\\Otvet:\ x=1\frac{3}{4}=1,75$