Đáp án:
`a,`
`x + 2x + 3x + 4x + .... + 2021x = 2012 × 2013`
`⇔ x × [1 + 2 + 3 + 4 + .... + 2021] = 2012 × 2013`
`⇔ x × [(2021 + 1) × 2021 ÷ 2] = 2012 × 2013`
`⇔ x × 2012 × 2021 ÷ 2=2012 × 2013`
`⇔ x × 2021 ÷ 2 = 2012 × 2013 ÷ 2012`
`⇔ x × 2021 ÷ 2 = 2013`
`⇔ x × 2021 = 2013 × 2`
`⇔ x × 2021 = 4026`
`⇔ x = 4026 ÷ 2021`
`⇔ x = 4026/2021`
Vậy `x = 4026/2021`
$\\$
`b,`
`(x - 1)/2021 + (x - 2)/2010 - (x - 3)/2009 = (x - 4)/2008`
`⇔ (x - 1)/2021 + (x - 2)/2010 = (x -4)/2008 + (x - 3)/2009`
`⇔ ( (x - 1)/2021 - 1) + ( (x - 2)/2010 - 1) = ( (x - 4)/2008 - 1) + ( (x - 3)/2009 - 1)`
`⇔ ( (x - 1) - 2021/2021) + ( (x - 2)/2010 - 2010/2010) = ( (x - 4)/2008 - 2008/2008) + ( (x - 3)/2009 - 2009/2009)`
`⇔ (x - 2012)/2021 + (x - 2012)/2010 = (x - 2012)/2008 + (x - 2012)/2009`
`⇔ (x - 2012)/2021 + (x - 2012)/2010 - (x - 2012)/2008 - (x - 2012)/2009=0`
`⇔ (x - 2012) × (1/2021 + 1/2010 - 1/2008 - 1/2009) = 0`
`⇔ x - 2012 = 0` (Vì `1/2021 + 1/2010 - 1/2008 - 1/2009 \ne 0`)
`⇔ x = 0 + 2012`
`⇔ x = 2012`
Vậy `x = 2012`
