Đáp án + Giải thích các bước giải:

Bài `1`: Tìm `x`, biết:

 `a_1 ,`

`5 4/7 : x = 13`

`=> 39/7 : x = 13`

`=> x = 39/7 : 13`

`=> x = 3/7`

Vậy `x = 3/7`

`a_2 ,`

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

`1/4 x = 2/3 - 3/2`

`1/4 x = -5/6`

` x = (-5)/6 : 1/4`

` x = -10/3`

Vậy `x = -10/3`

`b,`

`1/2 x + 2/3 x - 1 = -3 1/3`

`=> x. (1/2 + 2/3) - 1 = -10/3`

`=> x . 7/6 = (-10)/3 + 1`

`=> x. 7/6 = -7/3`

`=> x = -7/3 : 7/6`

`=> x = -2`

Vậy `x = -2`

`c,`

`x^2 - x = 0`

`=> x . (x - 1) = 0`

`<=> x = 0` hoặc `x - 1 = 0`

`<=> x = 0` hoặc `x = 1`

Vậy `x = {0; 1}`

`d,`

`5/3 x - 2/3 x = 1`

`x. (5/3 - 2/3) = 1`

`x. 1 = 1`

`x = 1`

Vậy `x = 1`

`e,`

`4. (x - 1) - 2. (x - 2) = 3`

`=> 4x - 4 - 2x + 4 = 3`

`=> (4x - 2x) + (4 - 4) = 3`

`=> 2x = 3`

`=> x = 3/2`

Vậy `x = 3/2`

Bài `2`: Tìm `x`, biết

`a,`

`-3 + |x + 2| = 5`

`=> -3 + |x + 2| + 3 = 5 + 3`

`=> |x + 2| = 8`

`<=> x + 2 = 8` hoặc `x + 2 = -8`

`<=> x = 6` hoặc `x = -10`

Vậy `x = {6; -10}`

`b,`

`|x - 1| = 2022`

\(\left[ \begin{array}{l}x-1=2022\\x-1=-2022\end{array} \right.\) => \(\left[ \begin{array}{l}x=2023\\x=-2021\end{array} \right.\) 

Vậy `x = {2023; -2021}`

`c,`

`|4x - 2x| + (-2) = 2020`

`=>|x. (4 - 2)| = 2020 + 2`

`=>|x. 2| = 2022`

`<=> 2x = 2022` hoặc `2x = -2022`

`<=> x = 1011` hoặc `x = -1011` 

Vậy `x = ±1011`

`d,`

` x/(-20) = (-5)/x`

`=> 2x = (-20) . (-5)` (bc nầy là nhân chéo đấy nhá)

`=> 2x = 100`

`=> x = 50`

Vậy `x = 50`

`text{ chỗ nèo k rõ cứ bẻo nhá }`

$#hackprovip$

a1) `5 4/7:x=13`

`=> 39/7:x=13`

`=> x=39/7:13`

`=> x=39/7xx1/13`

`=> x=39/91`

`=> x=3/7`

Vậy `x=3/7`

a2) `3/2+1/4x=2/3`

`=> 3/2+x/4=2/3`

`=> x/4=2/3-3/2`

`=> x/4=-5/6`

`=> x=-5/6xx4`

`=> x=-20/6`

`=> x=-10/3`

Vậy `x=-10/3`

b) `1/2x+2/3x-1=-3 1/3`

`=> 1/2x+2/3x-1=-10/3`

`=> x/2+(2x)/(3)-1=-10/3`

`=> (7x)/(6)-1=-10/3`

`=> (7x)/(6)=-7/3`

`=> 7x=-7/3xx6`

`=> 7x=-42/3`

`=> 7x=-14`

`=> x=-14/7`

`=> x=-2`

Vậy `x=-2`

c) `x^2-x=0`

`=> x(x-1)=0`

TH1:

`x=0`

TH2:

`x-1=0`

`=> x=1`

Vậy `x∈{0;1}`

d) `5/3x-2/3x=1`

`=> (5x)/(3)-(2x)/(3)=1`

`=> (3x)/(3)=1`

`=> 3x=1xx3`

`=> 3x=3`

`=> x=1`

Vậy `x=1`

e) `4(x-1)-2(x-2)=3`

`=> 4x-4-2x-4=3`

`=> 2x=3`

`=> x=3/2`

Vậy `x=3/2`

Bài 8:

a) `-3+|x+2|=5`

`=> |x+2|=5-(-3)`

`=> |x+2|=8`

`=>` \(\left[ \begin{array}{l}x+2=8\\x+2=-8\end{array} \right.\) 
`=>` \(\left[ \begin{array}{l}x=8-2\\x=-8-2\end{array} \right.\)
`=>`  \(\left[ \begin{array}{l}x=6\\x=-10\end{array} \right.\) 

Vậy `x∈{6; -10}`

b)  `|x-1|=2022`

`=>` \(\left[ \begin{array}{l}x-1=2022\\x-1=-2022\end{array} \right.\)
`=>`  \(\left[ \begin{array}{l}x=2022+1\\x=-2022+1\end{array} \right.\)
`=>`  \(\left[ \begin{array}{l}x=2023\\x=-2021\end{array} \right.\) 

Vậy `x∈{2023; -2021}`

c) `|4-2x|+(-2)=2020`

`=> |4-2x|-2=2020`

`=> |4-2x|=2022`

`=>` \(\left[ \begin{array}{l}4-2x=2022\\4-2x=-2022\end{array} \right.\)
`=>`  \(\left[ \begin{array}{l}2x=-2018\\2x=2026\end{array} \right.\)
`=>`  \(\left[ \begin{array}{l}x=-1009\\x=1013\end{array} \right.\) 

Vậy `x∈{-1009;  1013}`

d) `(x)/(-20)=(-5)/(x)`

`=> x*x=-20*(-5)`

`=> x^2=100`

`=>  x=±10`

Vậy `x∈{10; -10}`