thay các chữ số bằng số a) bccb-abc+ab b) abc = dad :5 c) ( a+b+c) x 11=abc 02/10/2021 Bởi Arianna thay các chữ số bằng số a) bccb-abc+ab b) abc = dad :5 c) ( a+b+c) x 11=abc
`a)bccb-abc=ab` `<=>abc+ab=bccba×100+bx10+c+a×10+b` `<=>b×10+cx100+cx10+ba×110-b×990` `<=>c×109110×(a-b)×9` `<=>c×109a` `=>abc=910` \(\left[ \begin{array}{l}a=9\\b=1\\c=0\end{array} \right.\) `b)abc=dad:5` `<=>dad=abc×5` `<=>abc×5=dad=>a=1;d=5=>dad=515` `<=>1bc×51bc=103` `c)(a+b+c)×11=abc` `<=>abc:11=a+b+c` `<=>a×100+b×10+c=11×a+11×b+11×c` `<=>89×a=b+10×c` `b;c≥9=>a=1` `=>89=b×10+c` Hay `b=89-10×c=>c=8` `=>b=89-10×8=9` Vậy `abc=198` Bình luận
`a)bccb-abc=ab`
`<=>abc+ab=bccba×100+bx10+c+a×10+b`
`<=>b×10+cx100+cx10+ba×110-b×990`
`<=>c×109110×(a-b)×9`
`<=>c×109a`
`=>abc=910`
\(\left[ \begin{array}{l}a=9\\b=1\\c=0\end{array} \right.\)
`b)abc=dad:5`
`<=>dad=abc×5`
`<=>abc×5=dad=>a=1;d=5=>dad=515`
`<=>1bc×51bc=103`
`c)(a+b+c)×11=abc`
`<=>abc:11=a+b+c`
`<=>a×100+b×10+c=11×a+11×b+11×c`
`<=>89×a=b+10×c`
`b;c≥9=>a=1`
`=>89=b×10+c`
Hay `b=89-10×c=>c=8`
`=>b=89-10×8=9`
Vậy `abc=198`