`1.`

`a) A= (1/3 - 8/15 - 1/7) + [2/3 + (-7/15) + 8/7]`

`A= 1/3 - 8/15 - 1/7 + 2/3  -7/15 + 8/7`

`A=(1/3+2/3)-(8/15+7/15)+(8/7-1/7)`

`A= 3/3 - 15/15 + 7/7`

`A= 1 - 1 + 1`

`A= 1.`

`b) B= 0,25 + 3/5 -(1/8 - 2/5 + 5/4)`

`B=1/4 + 3/5 - 1/8 + 2/5 -5/4`

`B=(1/4-5/4)+(2/5+3/5) - 1/8`

`B= -4/4 + 5/5 -1/8`

`B=-1+1-1/8`

`B=0-1/8`

`B=-1/8.`

Nếu đề là: 

`b) B= 0.25 + 3/5 -(1/8 - 2/5 + 5/4)`

`B=0 + 3/5 - 1/8 + 2/5 -5/4`

`B=-(1/8+5/4)+(2/5+3/5)`

`B=-(1/8+10/8)+5/5`

`B=-11/8+1`

`B=-11/8+8/8`

`B=-3/8.`

`2.`

`a) x^x-1=16`

`b)(x-2)^2 = 25`

`⇔(x-2)^2=(-5)^2=5^2`

`⇒` \(\left[ \begin{array}{l}x-2=5\\x-2=-5\end{array} \right.\) 

`⇒` \(\left[ \begin{array}{l}x=5+2\\x=-5+2\end{array} \right.\) 

`⇒` \(\left[ \begin{array}{l}x=7\\x=-3\end{array} \right.\) 

Vậy `x∈{-3;7}.`

Bạn viết lại đề bài 2a nha 

Bài $1$.

$a$) `A = (1/3 - 8/{15} - 1/7) + (2/3 + {-7}/{15} + 8/7)`

`⇔ A = 1/3 - 8/{15} - 1/7 + 2/3 - 7/{15} + 8/7`

`⇔ A = 1 - 1 + 1`

`⇔ A = 1`

$b$) `B = 0,25 + 3/5 - (1/8 - 2/5 + 5/4)`

`⇔ B = 1/4 + 3/5 - 1/8 + 2/5 - 5/4`

`⇔ B = -1 + 1 - 1/8`

`⇔ B = -1/8`

Bài $2$.

$b$) $(x-2)^2 = 25$

$⇔ x-2 = ± 5$

$⇒ x$ $∈$ `{-3;7}`

  Vậy $x$ $∈$ `{-3;7}`