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\(\overline{12abc0}=\overline{abc}\cdot80\)
\(\overline{12abc}=\overline{abc}\cdot8\)
abc lớn nhất bằng 999. Khi đó \(\overline{abc}\cdot8=999\cdot8=7992< \overline{12abc}\)
Suy ra không có giá trị phù hợp để biểu thức trên đúng.
![](https://rs.olm.vn/images/avt/0.png?1311)
Ta có : \(12abc0=810\times abc\)
\(120000+abc\times10=810\times abc\)
\(120000=810\times abc-abc\times10\)
\(120000=abc\times\left(810-10\right)\)
\(120000=abc\times800\)
\(120000:800=abc\)
\(abc=150\)
Vậy \(abc=150\)
Học tốt
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12:
\(B=\left(1-\dfrac{1}{2}\right)\left(1-\dfrac{1}{3}\right)\left(1-\dfrac{1}{4}\right)\left(1-\dfrac{1}{5}\right)...\left(1-\dfrac{1}{2004}\right)\)
\(B=\left(\dfrac{2}{2}-\dfrac{1}{2}\right)\left(\dfrac{3}{3}-\dfrac{1}{3}\right)\left(\dfrac{4}{4}-\dfrac{1}{4}\right)...\left(\dfrac{2004}{2004}-\dfrac{1}{2004}\right)\)
\(B=\dfrac{1}{2}\cdot\dfrac{2}{3}\cdot\dfrac{3}{4}\cdot\dfrac{4}{5}...\cdot\dfrac{2003}{2004}\)
\(B=\dfrac{1\cdot2\cdot3\cdot...\cdot2002\cdot2003}{2\cdot3\cdot4\cdot...\cdot2003\cdot2004}\)
\(B=\dfrac{1}{2004}\)
`12) B=(1-1/2)(1-1/3)(1-1/4)(1-1/5)....(1-1/2003)(1-1/2004)`
`B=1/2 . 2/3 . 3/4 . 4/5 .... 2002/2003 . 2003/2004`
`B= [1.2.3.4......2002.2003]/[2.3.4.5.....2003.2004]`
`B=1/2004`
`7)`
`a)[254xx399-145]/[254+399xx253]`
`=[(253+1)xx399-145]/[254+399xx253]`
`=[253xx399+399-145]/[254+399xx253]`
`=[254+253xx399]/[254+253xx399]=1`
`b)[5932+6001xx5931]/[5932xx6001-69]`
`=[5932+6001xx5931]/[(5931+1)xx6001-69]`
`=[5932+6001xx5931]/[5931xx6001+6001-69]`
`=[5932+6001xx5931]/[5932+6001xx5931]=1`
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