`bb5)`

`x/3 - ( 2x + 1 )/6 = x/6 - x`

`<=> ( 2x )/6 - ( 2x+1 )/6 = x/6 - ( 6x )/6`

`=> 2x-( 2x+1) = x - 6x`

`<=> 2x-2x-1=-5x`

`<=> -1=-5x`

`<=> 1=5x`

`<=> x = 1/5``(tm)`

Vậy `S={1/5}` 

`5)x/3-[2x+1]/6=x/6-x`

`<=>[2x-2x-1]/6=[x-6x]/6`

`<=>2x-2x-1=x-6x`

`<=>2x-2x-x+6x=1`

`<=>5x=1`

`<=>x=1/5`

Vậy `S={1/5}`

____________________________________

`6)[x+1]/2004+[x+2]/2003=[x+3]/2002+[x+4]/2001`

`<=>[x+1]/2004+1+[x+2]/2003+1=[x+3]/2002+1+[x+4]/2001+1`

`<=>[x+1+2004]/2004+[x+2+2003]/2003-[x+3+2002]/2002-[x+4+2001]/2001=0`

`<=>[x+2005]/2004+[x+2005]/2003-[x+2005]/2002-[x+2005]/2001=0`

`<=>(x+2005)(1/2004+1/2003-1/2002-1/2001)=0`

     Mà `1/2004+1/2003-1/2002-1/2001 \ne 0`

  `=>x+2005=0`

`<=>x=-2005`

Vậy `S={-2005}`

\(\dfrac{x}{3}-\dfrac{2x+1}{6}=\dfrac{x}{6}-x\)

\(\Leftrightarrow\dfrac{2x-\left(2x+1\right)}{6}=\dfrac{x-6x}{6}\)

\(\Leftrightarrow2x-2x-1=x-6x\)

\(\Leftrightarrow-5x=-1\)

\(\Leftrightarrow x=\dfrac{1}{5}\)

Vậy \(S=\left\{\dfrac{1}{5}\right\}\)

bon gà

+) Ta có: \(15-x+17=-\left(-6\right)+\left|-12\right|\)

        \(\Leftrightarrow-x+32=6+12\)

        \(\Leftrightarrow-x=18-32\)

        \(\Leftrightarrow x=14\)

Vậy......

+) Ta có: \(x+20=-\left(-23\right)\)

        \(\Leftrightarrow x+20=23\)

        \(\Leftrightarrow x=3\)

Vậy........

+) Ta có: \(-\left|-5\right|-\left(-x\right)+4=3-\left(-25\right)\)

       \(\Leftrightarrow-5+x+4=3+25\)

       \(\Leftrightarrow x=29\)

Vậy.....

Chúc e hok tốt và giữ lời hứa nhá ^_^

5A

1B

3C

4D

\(#TyHM\)

VIII

1 Have you ever been

2 Did you go

3 went

4 got up

5 have never got 

6 got 

7 Have you ever seen

D

I

1 B

2 A

3 D

4 A

5 C