Câu hỏi của Phan Công Bằng

Question

Tìm $abc$ biết $1:0,abc=a+b+c$

Nguyen Kha Vy · Accepted Answer

Ta có :$1:0,abc=a+b+c$

$\Rightarrow1:\dfrac{abc}{1000}=a+b+c$

$\Rightarrow1000:abc=a+b+c$

$\Rightarrow abc\times\left(a+b+c\right)=1000$

Mà $1000=500\times2=100\times10=250\times4=200\times5=125\times8$

$\Rightarrow abc\times\left(a+b+c\right)=125\times8=125\times\left(1+2+5\right)$

Vậy abc =125Câu trả lời: Ta có :$1:0,abc=a+b+c$

$\Rightarrow1:\dfrac{abc}{1000}=a+b+c$

$\Rightarrow1000:abc=a+b+c$

$\Rightarrow abc\times\left(a+b+c\right)=1000$

Mà $1000=500\times2=100\times10=250\times4=200\times5=125\times8$

$\Rightarrow abc\times\left(a+b+c\right)=125\times8=125\times\left(1+2+5\right)$

Vậy abc =125

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