`a)`
`1/2x+3/5x=-33/25`
`\to (1/2+3/5)x=-33/25`
`\to 11/10x=-33/25`
`\to x=-33/25:11/10`
`\to x=-6/5`
Vậy `x=-6/5`
``
`b)`
`(2/3x-4/9)(1/2+(-3)/7:x)=0`
Trường hợp 1:
`2/3x-4/9=0`
`\to 2/3x=4/9`
`\to x=4/9:2/3`
`\to x=2/3`
Trường hợp 2:
`1/2+(-3)/7:x=0`
`\to -3/7:x=-1/2`
`\to x=-3/7 : (-1/2)`
`\to x=6/7`
Vậy `x\in{2/3;6/7}`
``
`c)`
`(x+5)/(2005)+(x+6)/(2004)+(x+7)/(2003)=-3`
`\to (x+5)/(2005)+(x+6)/(2004)+(x+7)/(2003)+3=0`
`\to ((x+5)/(2005)+1)+((x+6)/(2004)+1)+((x+7)/(2003)+1)=0`
`\to ((x+5)/(2005)+2005/2005)+((x+6)/(2004)+2004/2004)+((x+7)/(2003)+2003/2003)=0`
`\to (x+5+2005)/2005+(x+6+2004)/2004+(x+7+2003)/(2003)=0`
`\to (x+2010)/2005+(x+2010)/2004+(x+2010)/2003=0`
`\to (x+2010)(1/2005+1/2004+1/2003)=0`
`\to x+2010=0` (vì `1/2005+1/2004+1/2003\ne0`)
`\to x=-2010`
Vậy `x=-2010`
