`1.`

`a)  -2/3.x+4=-12`

` ⇔ -2/3.x =-12-4`

` ⇔ -2/3.x =-16`

` ⇔ x =-16 : (-2/3)`

` ⇔ x =-16 . (-3/2)`

` ⇔ x = -24.`

Vậy `  x = -24.`

`b) -3/4 + 1/4 : x =-3`

`⇔-3/4 + 1/4 . 1/x =-3`

`⇔ 1/4. 1/x =-3+3/4`

`⇔ 1/4. 1/x =-9/4`

`⇔1/x =-9/4:1/4`

`⇔1/x =-9`

`⇔x =1:-9`

`⇔x =-1/-9`

Vậy ` x =-1/-9.`

 `c) \frac{x+1}{2} + \frac{x+1}{3} + \frac{x+1}{4} = \frac{x+1}{5}`

 `⇔ \frac{x+1}{2} + \frac{x+1}{3} + \frac{x+1}{4} - \frac{x+1}{5}=0`

 `⇔ (x+1)(1/2+1/3+1/4-1/5)=0` 

 `⇔ x+1=0`  ( vì `1/2+1/3+1/4-1/5\ne0` )

`⇔ x=-1.`

Vậy `x=-1.`

`2.`

`a) 2^1. 2^2. 2^3. 2^4. 2^x=1024`

`⇔2^{1+2+3+4+x}=1024`

`⇔2^{10+x}=2^10`

`⇒10+x=10`

`⇒x=0.`

Vậy `x=0.`

`b)\frac{x+1}{2009}+\frac{x+21}{2008}+\frac{x+3}{2007}=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}` 

`⇔\frac{x+1}{2009}+\frac{x+21}{2008}+\frac{x+3}{2007}+3=\frac{x+10}{2000}+\frac{x+11}{1999}+\frac{x+12}{1998}+3` 

`⇔\frac{x+1}{2009}+1+\frac{x+21}{2008}+1+\frac{x+3}{2007}+1=\frac{x+10}{2000}+1+\frac{x+11}{1999}+1+\frac{x+12}{1998}+1` 

`⇔\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}=\frac{x+2010}{2000}+\frac{x+2010}{1999}+\frac{x+2010}{1998}` 

`⇔\frac{x+2010}{2009}+\frac{x+2010}{2008}+\frac{x+2010}{2007}-\frac{x+2010}{2000}-\frac{x+2010}{1999}-\frac{x+2010}{1998}=0` 

`⇔(x+2010)(1/2009+1/2008+1/2007-1/2000-1/1999-1/1998)=0`

`⇔x+2010=0` ( vì `1/2009+1/2008+1/2007-1/2000-1/1999-1/1998`)`\ne0` 

`⇔x=0-2010=-2010.`

Vậy `x=-2010.`

`3.`

`a)|x-5|=3`
`+) x-5 \ge 0 ⇔ x \ge  5⇒ |x-5| = x-5 `
`+) x-5 < 0 ⇔ x <  5 ⇒ |x-5| = 5-x `

Xét trường hợp: `x \ge  5 `

`⇒  x-5 = 3`

`⇒  x=3+5 = 8 ` (TMĐK)

Xét trường hợp: ` x <  5  `

`⇒  5-x = 3`

`⇒  x=5-3 = 2 ` (TMĐK)

Vậy tập nghiệp của phương trình: `S={8,2}.`

`b)|2x-1|=1-3/4`

`⇔|2x-1|=1/4`
`+) 2x-1\ge 0⇔ x \ge  1/2 ⇒ |2x-1| = 2x-1 `
`+) 2x-1< 0⇔ x <  1/2 ⇒ |2x-1| = 1-2x`

Xét trường hợp: ` x \ge  1/2 `

`⇒  2x-1 = 1/4`

`⇒  x=5/8 ` (TMĐK)

Xét trường hợp: ` x <  1/2 `

`⇒ 1-2x = 1/4`

`⇒  x=3/8 ` (TMĐK)

Vậy tập nghiệp của phương trình: `S={5/8,3/8}.`

`c) |4x+2|-3=1/7`

`⇔ |4x+2|=1/7+3`

`⇔ |4x+2|=22/7`

`+) 4x+2\ge 0⇔ x \ge  -1/2 ⇒ |4x+2| = 4x+2 `
`+) 4x+2< 0⇔ x <  -1/2 ⇒ |4x+2| = -4x-2 `

Xét trường hợp: ` x \ge  -1/2 `

`⇒  4x+2 = 22/7`

`⇒  x=2/7 ` (TMĐK)

Xét trường hợp: ` x <  -1/2 `

`⇒  -4x-2 = 22/7`

`⇒  x=-9/7 ` (TMĐK)

Vậy tập nghiệp của phương trình: `S={2/7,-9/7}.`

Đáp án:1)a)x=24

b)x=-1/9

 2)a)x=0

b)x=-2010

Giải thích các bước giải:

 1)a)-2/3x+4=-12

<=>-2/3x=-16

<=>x=24

b)-3/4+1/4/x=-3

<=>1/4/x=-9/4

<=>x=1/4/-9/4

<=>x=-1/9

2)a)2^1*2^2*2^3*2^4*2^x=1024

<=>2^(1+2+3+4)*2^x=1024

<=>2^10*2^x=2^10

<=>2^x=2^10/2^10

<=>2^x=1

<=>2^x=2^0

<=>x=0

b)(x+1)/2009+(x+2)/2008+(x+3)/2007=(x+10)/2000+(x+11)/1999+(x+12)/1998

<=>1+(x+1)/2009+1+(x+2)/2008+1+(x+3)/2007=1+(x+10)/2000+1+(x+11)/1999+1+(x+12)/1998

<=>(x+1+2009)/2009+(x+2+2008)/2008+(x+3+2007)/2007=(x+10+2000)/2000+(x+11+1999)/1999+(x+12+1998)/1998

<=>(x+2010)/2009+(x+2010)/2008+(x+2010)/2007=(x+2010)/2000+(x+2010)/1999+(x+2010)/1998

<=>(x+2010)/2009+(x+2010)/2008+(x+2010)/2007-(x+2010)/2000+(x+2010)/1999+(x+2010)/1998=0

<=>(x+2010)(1/2009+1/2008+1/2007-1/2000-1/1999-1/1998)=0

vì 1/2009+1/2008+1/2007-1/2000-1/1999-1/1998 khác 0

nên x+2010=0

<=>x=-2010