Question

помогите решить. нужно перевести в дробь. там типа: 1.5(2)=
$\frac{152 - 5}{90}$
и т.д. я просто не очень понимаю как​

Answer 1

$1,(23)=\frac{123-1}{99}=\frac{122}{99}=1\frac{23}{99}$

$1,(14)=\frac{114-1}{99}=\frac{113}{99}=1\frac{14}{99}$

$2,(16)=2\frac{16-0}{99}=2\frac{16}{99}$

$1,5(2)=\frac{152-15}{90}=\frac{137}{90}=1\frac{47}{90}$

$1,3(4)=1\frac{34-3}{90}=1\frac{31}{90}$

$2,4(3)=2\frac{43-4}{90}=2\frac{39}{90}$

$1,5(23)=1\frac{523-5}{990}=1\frac{518}{990}=1\frac{259}{495}$

$2,4(13)=2\frac{413-4}{990}=2\frac{409}{990}$

$1,3(14)=1\frac{314-3}{990}=1\frac{311}{990}$

$1,5(231)=1\frac{5231-5}{9990}=1\frac{5226}{9990}=1\frac{871}{1665}$

$2,1(425)=2\frac{1425-1}{9990}=2\frac{1424}{9990}=2\frac{712}{4995}$

$2,1(213)=2\frac{1213-1}{9990}=2\frac{1212}{9990}=2\frac{202}{1665}$

$1,52(23)=1\frac{5223-52}{9900}=1\frac{5171}{9900}$

$2,12(24)=2\frac{1224-12}{9900}=2\frac{1212}{9900}=2\frac{101}{825}$

$2,12(13)=2\frac{1213-12}{9900}=2\frac{1201}{9900}$

Answer 2

$1)\; \; a=1,(23)=1,232323...\\\\100a=123,232323....=123,(23)\\\\100a-a=123,(23)-1,(23)\\\\99a=122\; \; ,\; \; a=\frac{122}{99}\\\\2)\; \; a=1,5(2)=1,522222...\\\\10a=15,(2)=15,222222...\\\\100a=152,222222...=152,(2)\\\\100a-10a=152,(2)-15,(2)\\\\90a=137\; \; ,\; \; a=\frac{137}{90}\\\\3)\; \; a=1,5(23)=1,5232323...\\\\10a=15,(23)=15,232323...\\\\1000a=1523,(23)=1523,232323...\\\\1000a-10a=1523,(23)-15,(23)\\\\990a=1508\; \; ,\; \; \; a=\frac{1508}{990}$

$4)\; \; a=1,5(231)\\\\10a=15,(231)\\\\10000a=15231,(231)\\\\10000a-10a=15231,(231)-15,(231)\\\\9990a=15216\; \; ,\; \; \; a=\frac{15216}{9990}\\\\5)\; \; a=1,52(23)\\\\100a=152,(23)\\\\10000a=15223,(23)\\\\10000a-100a=15223,(23)-152,(23)\\\\9900a=15071\; \; ,\; \; a=\frac{15071}{9900}$