Lm câu a thoii đc chớ ak?

a,

`|4x-4|+|x-2|=5`

`=> 4|x-1|+|x-2|=5` (1)

Ker bảng :

x           |            1                           2            |

x - 1      |     -      0             +           |      +     |

x - 2     |      -      |              -            0      +     |
Th1 : x < 1

=> (1) `⇔ 4 (1-x)+(2-x)=5`

`⇔4-4x+2-x=5`

`⇔6-5x=5`

`⇔5x=1`

`⇔x=1/5` (tm x <1)

Th2 : `1≤x≤2`

`=> (1) ⇔ 4x-4+2-x=5`

`⇔3x-2=5`

`⇔3x=7`

`⇔x=7/3` (ktmđk)

Th2 : x >2

`=> (1) ⇔ 4x-4+x-2=5`

`=> 5x - 6 = 5`

`=> 5x = 11`

`=> x = 11/5` (tm)
Vậy `x∈{11/5 ; 1/5}`

 

Đáp án:

b. \(\left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x =  - \dfrac{{15}}{2}
\end{array} \right.\)

Giải thích các bước giải:

\(\begin{array}{l}
a.\left| {4x - 4} \right| + \left| {x - 2} \right| = 5\\
 \to \left[ \begin{array}{l}
4x - 4 + x - 2 = 5\left( {DK:x \ge 2} \right)\\
4x - 4 - x + 2 = 5\left( {DK:2 > x \ge 1} \right)\\
4x - 4 + x - 2 =  - 5\left( {DK:1 > x} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
5x = 11\\
3x = 7\\
5x = 1
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{{11}}{5}\left( {TM} \right)\\
x = \dfrac{7}{3}\left( l \right)\\
x = \dfrac{1}{5}\left( {TM} \right)
\end{array} \right.\\
b.\left| {x + 3} \right| + \left| {x + 4} \right| = 8\\
 \to \left[ \begin{array}{l}
x + 3 + x + 4 = 8\left( {DK:x \ge  - 3} \right)\\
 - x - 3 + x + 4 = 8\left( {DK: - 3 > x \ge  - 4} \right)\\
x + 3 + x + 4 =  - 8\left( {DK: - 4 > x} \right)
\end{array} \right.\\
 \to \left[ \begin{array}{l}
2x = 1\\
1 = 8\left( l \right)\\
2x =  - 15
\end{array} \right. \to \left[ \begin{array}{l}
x = \dfrac{1}{2}\\
x =  - \dfrac{{15}}{2}
\end{array} \right.\left( {TM} \right)
\end{array}\)