Tìm x biết: $x-\dfrac{8-\dfrac{8}{5}+\dfrac{8}{25}-\dfrac{8}{125}}{9-\dfrac{9}{5}+\dfrac{9}{25}-\dfrac{9}{125}} : \dfrac{161616}{151515} = \dfrac{4+\dfrac{4}{73}-\dfrac{4}{115}}{5+\dfrac{5}{73}-\dfrac{1}{23}}$
Tìm x biết: $x-\dfrac{8-\dfrac{8}{5}+\dfrac{8}{25}-\dfrac{8}{125}}{9-\dfrac{9}{5}+\dfrac{9}{25}-\dfrac{9}{125}} : \dfrac{161616}{151515} = \dfrac{4+\dfrac{4}{73}-\dfrac{4}{115}}{5+\dfrac{5}{73}-\dfrac{1}{23}}$
Đáp án:
`x=49/30`
Giải thích các bước giải:
`x-{8-8/5+8/25-8/125}/{9-9/5+9/25-9/125}:(161616)/(151515)={4+4/73-4/115}/{5+5/73-1/23}`
`=>x-{8/1-8/5+8/25-8/125}/{9/1-9/5+9/25-9/125}:16/15={4/1+4/73-4/115}/{5/1+5/73-5/115}`
`=>x-{8(1/1-1/5+1/25-1/125)}/{9(1/1-1/5+1/25-1/125)}. 15/16={4(1/1+1/73-1/115)}/{5(1/1+1/73-1/115)}`
`=>x-8/9 . 15/16=4/5`
`=>x-1/3 . 5/2=4/5`
`=>x-5/6=4/5`
`=>x=4/5+5/6`
`=>x=24/30+25/30`
`=>x=49/30`
Vậy `x=49/30`
$\text{Đáp án + Giải thích các bước giải:}$
$x-\dfrac{8-\dfrac{8}{5}+\dfrac{8}{25}-\dfrac{8}{125}}{9-\dfrac{9}{5}+\dfrac{9}{25}-\dfrac{9}{125}} : \dfrac{161616}{151515} = \dfrac{4+\dfrac{4}{73}-\dfrac{4}{115}}{5+\dfrac{5}{73}-\dfrac{1}{23}}$
$x-\dfrac{8-\dfrac{8}{5}+\dfrac{8}{25}-\dfrac{8}{125}}{9-\dfrac{9}{5}+\dfrac{9}{25}-\dfrac{9}{125}} : \dfrac{161616}{151515} = \dfrac{4+\dfrac{4}{73}-\dfrac{4}{115}}{5+\dfrac{5}{73}-\dfrac{5}{115}}$
$⇔x-\dfrac{8(1-\dfrac{1}{5}+\dfrac{1}{25}-\dfrac{1}{125})}{9(1-\dfrac{1}{5}+\dfrac{1}{25}-\dfrac{1}{125})}:\dfrac{16}{15}=\dfrac{4(1+\dfrac{1}{73}-\dfrac{1}{115})}{5(1+\dfrac{1}{73}-\dfrac{1}{115})}$
$⇔x-\dfrac{8}{9}.\dfrac{15}{16}=\dfrac{4}{5}$
$⇔x-\dfrac{5}{6}=\dfrac{4}{5}$
$⇔x=\dfrac{4}{5}+\dfrac{5}{6}$
$⇔x=\dfrac{24}{30}+\dfrac{25}{30}$
$⇔x=\dfrac{49}{30}$
Vậy $x=\dfrac{49}{30}$
$#minosuke$