c) l x - 5 l = 2x

\(\Leftrightarrow\orbr{\begin{cases}x-5=2x\\x-5=-2x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x-2x=5\\x+2x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}-x=5\\3x=5\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\x=\frac{5}{3}\end{cases}}\)

Hok tốt!!!!!!!

Tìm x, biết:

a) |2x + 1| = 17

<=>\(\orbr{\begin{cases}2x+1=17\\2x+1=-17\end{cases}}\) 

<=>\(\orbr{\begin{cases}2x=16\\2x=-18\end{cases}}\)

<=> \(\hept{\begin{cases}x=8\\x=-9\end{cases}}\)

Phần d không biết , thông cảm

a) Ta có : 

| 2x + 1 | = 17

\(\Rightarrow\orbr{\begin{cases}2x+1=17\\2x+1=-17\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}2x=16\\2x=-18\end{cases}}\Rightarrow\orbr{\begin{cases}x=8\\x=-9\end{cases}}\)

b) | 3x - 17 | = | 2x - 3 |

Xét các trường hợp :

+) 3x - 17 = - 2x + 3

=> 3x + 2x = 3 + 17

=> 5x = 20

=> x = 4

+) 3x - 17 = 2x - 3

=> 3x - 2x = -3 + 17

=> x = 14

+) -3x + 17 = 2x - 3

=> 2x + 3x = 17 + 3x

=> 5x = 20

=> x = 4

c) | x - 5 | = 2x

\(\Rightarrow\orbr{\begin{cases}x-5=2x\\x-5=-2x\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}-x=5\\3x=5\end{cases}}\Rightarrow\orbr{\begin{cases}x=5\\x=\frac{5}{3}\end{cases}}\)

a) \(\dfrac{x}{3}=\dfrac{4}{12}\Rightarrow x=\dfrac{4}{12}\cdot3=\dfrac{12}{12}=1\)

b) \(\dfrac{x-1}{x-2}=\dfrac{3}{5}\) (Điều kiện : \(x\ne2\))

\(\Rightarrow5\left(x-1\right)=3\left(x-2\right)\)

\(\Leftrightarrow5x-5=3x-6\Leftrightarrow5x-3x=-6+5\Leftrightarrow2x=-1\Leftrightarrow x=-\dfrac{1}{2}\)

c) \(2x:6=\dfrac{1}{4}\Leftrightarrow2x=\dfrac{1}{4}\cdot6=\dfrac{6}{4}=\dfrac{3}{2}\Leftrightarrow x=\dfrac{3}{2}:2=\dfrac{3}{2}\cdot\dfrac{1}{2}=\dfrac{3}{4}\)

d) \(\dfrac{x^2+x}{2x^2+1}=\dfrac{1}{2}\)

\(\Rightarrow2\left(x^2+x\right)=2x^2+1\)

\(\Leftrightarrow2x^2+2x=2x^2+1\)

\(\Leftrightarrow2x^2+2x-2x^2=1\Leftrightarrow2x=1\Leftrightarrow x=\dfrac{1}{2}\).

a) 2x-5=3+2x-7x

2x-2x+7x=3+5

7x=8

  x=8/7

vậy x=8/7

a) 2x - 5 = 3 + 2x - 7x

=> 2x - 2x + 7x = 3 +5 

=> 7x = 8

=> x = 8/7

b) \(\left(2x-1\right)^2=\left(2x-1\right)^5\)

=> \(\left(2x-1\right)^2-\left(2x-1\right)^5=0\)

=> \(\left(2x-1\right)^2\left[1-\left(2x-1\right)^3\right]=0\)

=> \(\orbr{\begin{cases}\left(2x-1\right)^2=0\\1-\left(2x-1\right)^3=0\end{cases}}\)

=> \(\orbr{\begin{cases}2x-1=0\\\left(2x-1\right)^3=1\end{cases}}\)

=> \(\orbr{\begin{cases}2x=1\\2x-1=1\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\2x=2\end{cases}}\)

=> \(\orbr{\begin{cases}x=\frac{1}{2}\\x=1\end{cases}}\)