`text{BÔNG gởi pẹn <3}`

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Đáp án:

`a, x = -5/6`

`b, x = 1`

`c, x in { 2/5 ; -4/7}`

`d, x in { 7/3 ; -2/3}`

`e, x = 8/7`

`f, x in { 1/2 ; -3/2}`

Giải thích các bước giải:

 `a, x - 3/4 = 11/12 - 5/2`

`=> x - 3/4 = 11/12 - 30/12`

`=> x - 3/4 = -19/12`

`=> x = -19/12 + 3/4`

`=> x = -19/12 + 9/12`

`=> x = -10/12`

`=> x = -5/6`

`text{Vậy :}` `x = -5/6`

`b, -4/7 . x + 3/4 = 5/28`

`=> -4/7 . x = 5/28 - 3/4`

`=> -4/7 . x = 5/28 - 21/28`

`=> -4/7 . x = -16/28`

`=> -4/7 . x = -4/7`

`=> x = ( -4/7 ) : ( -4/7 )

`=> x = 1`

`text{Vậy :}` `x = 1`

`c, ( x -2/5 ) . ( x +4/7)=0`

`=>` \(\left[\begin{array}{l}x - \frac{2}{5} = 0\\x + \frac{4}{7} = 0\end{array} \right.\) `=>`\(\left[ \begin{array}{l}x = \frac{2}{5}\\x =- \frac{4}{7}\end{array} \right.\)

`text{Vậy :}` `x in { 2/5 ; -4/7}`

`d, -1 + | x -5/6| = 1/2`

`=> | x -5/6 | = 1/2 + 1`

`=> | x -5/6 | = 1/2 +2/2`

`=> | x -5/6 | = 3/2`

`=> x -5/6 =` `+-``3/2`

`=> x in { 7/3 ; -2/3}`

`text{Vậy :}` ` x in { 7/3 ; -2/3}`

`e, x + 5/8 . x - 12/16 . x = 1`

`=> x . ( 1 + 5/8 - 6/8 ) = 1`

`=> x . ( 8/8 + 5/8 - 6/8 ) = 1`

`=> x . 7/8 = 1`

`=> x = 1 : 7/8`

`=> x = 8/7`

`text{Vậy :}` ` x = 8/7`

`f, ( x + 1/2 ) ^2 -1/3 = 6/9`

`=> ( x + 1/2 )^2 = 6/9 + 1/3`

`=> ( x +1/2)^2 = 6/9 +3/9`

`=> x +1/2 = ` `+-` `1`

`=> x in { 1/2 ; -3/2}`

`text{Vậy :}` ` x in { 1/2 ; -3/2}`

Đáp án:

$\\$

`a,`

`x- 3/4 = 11/12 - 5/2`

`↔ x - 3/4 =(-19)/12`

`↔ x = (-19)/12 + 3/4`

`↔ x =(-5)/6`

Vậy `x=(-5)/6`

$\\$

`b,`

`(-4)/7x + 3/4 = 5/28`

`↔ (-4)/7x = 5/28 - 3/4`

`↔ (-4)/7x = (-4)/7`

`↔ x =(-4)/7 ÷ (-4)/7`

`↔ x = (-4)/7 × (-7)/4`

`↔ x=1`

Vậy `x=1`

$\\$

`c,`

`(x-2/5) (x + 4/7) = 0`

`↔` \(\left[ \begin{array}{l}x-\dfrac{2}{5}=0\\x+\dfrac{4}{7}=0\end{array} \right.\)

`↔`  \(\left[ \begin{array}{l}x=0+\dfrac{2}{5}\\x=0-\dfrac{4}{7}\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=\dfrac{2}{5}\\x=\dfrac{-4}{7}\end{array} \right.\) 

Vậy `x=2/5` hoặc `x=(-4)/7`

$\\4

`d,`

`-1+ |x - 5/6| = 1/2`

`↔ |x - 5/6| = 1/2 - (-1)`

`↔ |x-5/6|=3/2`

`↔` \(\left[ \begin{array}{l}x-\dfrac{5}{6}=\dfrac{3}{2}\\x-\dfrac{5}{6}=\dfrac{-3}{2}\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=\dfrac{3}{2}+\dfrac{5}{6}\\x=\dfrac{-3}{2}+\dfrac{5}{6}\end{array} \right.\) 

`↔` \(\left[ \begin{array}{l}x=\dfrac{7}{3}\\x=\dfrac{-2}{3}\end{array} \right.\) 

Vậy `x=7/3` hoặc `x=(-2)/3`

$\\$

`e,`

`x + 5/8x - 12/16x = 1`

`↔ (1 + 5/8 - 12/16) x = 1`

`↔7/8x = 1`

`↔ x = 1 ÷ 7/8`

`↔ x = 1 × 8/7`

`↔ x = 8/7`

Vậy `x=8/7`

$\\$

`f,`

`(x+1/2)^2 - 1/3 = 6/9`

`↔ (x+1/2)^2=6/9+1/3`

`↔ (x+1/2)^2=1`

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

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

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

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

Vậy `x=1/2` hoặc `x=(-3)/2`