$1$. 

$a$) `|x| - 1/2 = 3/4`

`⇔ |x| = 3/4 + 2/4`

`⇔ |x| = 5/4`

`⇒` ` x = ± 5/4`

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

`⇔ |x-1/2| = 1/4 + 3/4`

`⇔ |x-1/2| = 1`

`⇒` \(\left[ \begin{array}{l}x-\dfrac{1}{2}=1\\x-\dfrac{1}{2}=-1\end{array} \right.\) 

`⇒` \(\left[ \begin{array}{l}x=\dfrac{3}{2} \\x= -\dfrac{1}{2}\end{array} \right.\) 

$c$) `1/{20} - | 3/2 - x| = 1/{40}`

`⇔ |3/2 - x| = 1/{20} - 1/{40}`

`⇔ |3/2 - x| = 1/{40}`

`⇒` \(\left[ \begin{array}{l}\dfrac{3}{2}-x=\dfrac{1}{40}\\\dfrac{3}{2}-x=\dfrac{-1}{40}\end{array} \right.\) 

$⇒$ \(\left[ \begin{array}{l}x = \dfrac{59}{40} \\x=\dfrac{61}{40}\end{array} \right.\) 

$d$)  `(x+1/3)^2 + 4/9 = 8/9`

`⇔ (x+1/3)^2 = 8/9 - 4/9`

`⇔ (x+1/3)^2 = 4/9`

`⇔ (x+1/3)^2 = (± 2/3)^2`

`⇒` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-1\end{array} \right.\) 

$2$.

$a$) `A = 1/{1.2} + 1/{2.3} + 1/{3.4} + 1/{4.5} + 1/{5.6} + 1/{6.7} + 1/{7.8}`

`⇔ A = 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8`

`⇔ A = 1 - 1/8`

`⇔ A = 7/8`.

$b$) `B = 2/{1.3} + 2/{3.5} + ..... + 2/{11.13}`

`⇔ B = 1 - 1/3 + 1/3 - 1/5 + ..... + 1/11 - 1/13`

`⇔ B = 1 - 1/{13}`

`⇔ B = 12/13`.

Đáp án:

1.

a) `x in { -5/4 ; 5/4 }` 

b) `x in { -1/2 ; 3/2 }`

c) `x in { 59/40 ; 61/40 }`

d) `x in { -1 ; 1/3 }`

2.

a) `A = 7/8`

b) `B = 12/13`

Giải thích các bước giải:

1.

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

`to |x| = 3/4 + 1/2`

`to |x| = 5/4`

`to x = 5/4` or `x = -5/4`

Vậy `x in { -5/4 ; 5/4 }`

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

`to |x-1/2| = 1/4 + 3/4`

`to |x-1/2| = 1`

`to` \(\left[ \begin{array}{l}x-\dfrac{1}{2}=1\\x-\dfrac{1}{2}=-1\end{array} \right.\) `to` \(\left[ \begin{array}{l}x=\dfrac{3}{2} \\x= -\dfrac{1}{2}\end{array} \right.\) 

Vậy `x in { -1/2 ; 3/2 }`

c) `1/20 - | 3/2 - x | = 1/40`

`to | 3/2 - x | = 1/2 - 1/40`

`to | 3/2 - x | = 1/40`

`to ` \(\left[ \begin{array}{l}\dfrac{3}{2}-x=\dfrac{1}{40}\\\dfrac{3}{2}-x=-\dfrac{1}{40}\end{array} \right.\) `to `\(\left[ \begin{array}{l}x = \dfrac{59}{40} \\x=\dfrac{61}{40}\end{array} \right.\) 

Vậy `x in { 59/40 ; 61/40 }`

d) `(x+1/3)^2 + 4/9 = 8/9`

`to (x+1/3)^2 = 8/9 - 4/9`

`to (x+1/3)^2 = 4/9`

`to ` \(\left[ \begin{array}{l}x+\dfrac{1}{3}=\dfrac{2}{3}\\x+\dfrac{1}{3}=-\dfrac{2}{3}\end{array} \right.\)  `to` \(\left[ \begin{array}{l}x=\dfrac{1}{3}\\x=-1\end{array} \right.\) 

Vậy `x in { -1 ; 1/3 }`

$\\$

2.

a) `A = 1/1.2 + 1/2.3 + 1/3.4 + 1/4.5 + 1/5.6 + 1/6.7 + 1/7.8`

`= 1 - 1/2 + 1/2 - 1/3 + 1/3 - 1/4 + 1/4 - 1/5 + 1/5 - 1/6 + 1/6 - 1/7 + 1/7 - 1/8`

`= 1 - 1/8 = 7/8`

Vậy `A = 7/8`

b) `B = 2/1.3 + 2/3.5 + ... + 2/11.13`

`= 1 - 1/3 + 1/3 - 1/5 + ..... + 1/11 - 1/13`

`= 1 - 1/13 = 12/13`

Vậy `B = 12/13`