a) \(\dfrac{x-4}{15}=\dfrac{5}{3}\)

\(\Leftrightarrow x-4=15.\dfrac{5}{3}\)

\(\Leftrightarrow x-4=25\)

\(\Leftrightarrow x=29\) thỏa \(x\inℤ\)

b) \(\dfrac{x}{4}=\dfrac{18}{x+1}\left(x\ne-1\right)\)

\(\Leftrightarrow x\left(x+1\right)=18.4\)

\(\Leftrightarrow x\left(x+1\right)=72\)

vì \(72=8.9=\left(-8\right).\left(-9\right)\)

\(\Leftrightarrow x\in\left\{8;-9\right\}\left(x\inℤ\right)\)

c) \(2x+3⋮x+4\) \(\left(x\ne-4;x\inℤ\right)\)

\(\Leftrightarrow2x+3-2\left(x+4\right)⋮x+4\)

\(\Leftrightarrow2x+3-2x-8⋮x+4\)

\(\Leftrightarrow-5⋮x+4\)

\(\Leftrightarrow x+4\in\left\{-1;1;-5;5\right\}\)

\(\Leftrightarrow x\in\left\{-5;-3;-9;1\right\}\)

a) |2x-1|=5-x

\(\Leftrightarrow\orbr{\begin{cases}2x-1=5-x\\2x-1=-5+x\end{cases}}\)

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

b)|2x-1|>2     <=>\(\orbr{\begin{cases}2x-1>2\\2x-1< -2\end{cases}\Leftrightarrow}\orbr{\begin{cases}x>\frac{3}{2}\\x< \frac{-1}{2}\end{cases}}\)

c)\(\Leftrightarrow-5< 3x-7< 5\)   <=>2/3<x<4

24 - 16(x - 1/2) = 23

=> 16(x - 1/2) = 24 - 23

=> 16(x - 1/2) = 1

=> x - 1/2 = 1/16

=> x = 1/16 + 1/2

=> x = 9/16

\(24-16(x-\frac{1}{2})=23\)

\(16(x-\frac{1}{2})=24-23\)

\(16(x-\frac{1}{2})=1\)

\(x-\frac{1}{2}=\frac{1}{16}\)

\(x=\frac{1}{16}+\frac{1}{2}\)

\(x=\frac{9}{16}\)

Vậy số thực x cần tìm là \(\frac{9}{16}\)

Chúc bạn hok tốt ~

x - 2 = | 2x + 3 |

\(\Leftrightarrow\orbr{\begin{cases}x-2=2x+3\\x-2=-2x-3\end{cases}\Leftrightarrow\orbr{\begin{cases}-x=5\\3x=-1\end{cases}\Leftrightarrow}\orbr{\begin{cases}x=-5\\x=-\frac{1}{3}\end{cases}}}\)

có 2 th

x -2 = 2x + 3 suy ra x + 5 = 0 suy ra x = 5

-x + 2 = 2x+3 suy ra 3x + 1 = 0 suy ra x= -1/3

\(\Leftrightarrow\orbr{\begin{cases}2x+3=x-2\\2x+3=2-x\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}2x-x=-2-3\\2x+x=2-3\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=-5\\3x=-1\Rightarrow x=\frac{-1}{3}\end{cases}}\)

k mình nhé, cảm ơn bạn nhiều

Ta có : \(2x-5⋮x-3\)

\(\Rightarrow2x-6+1⋮x-3\)

\(\Rightarrow2\left(x-3\right)+1⋮x-3\)

Mà : \(2\left(x-3\right)⋮x-3\)suy ra : \(1⋮x-3\)

\(\Rightarrow x-3\inƯ\left(1\right)=\left\{-1;1\right\}\)

\(\Rightarrow x\in\left\{2;4\right\}\)

\(\left(2x-5\right)⋮\left(x-3\right)\)

\(\Rightarrow\left(2x-6+1\right)⋮\left(x-3\right)\)

Vì \(\left(2x-6\right)⋮\left(x-3\right)\)nên \(1⋮\left(x-3\right)\)

\(\Rightarrow x-3\inƯ\left(1\right)=\left\{\pm1\right\}\)

Trường hợp : \(x-3=-1\)

\(\Leftrightarrow x=-1+3\)

\(\Leftrightarrow x=2\)

Trường hợp : \(x-3=1\)

Vì \(\left|2x-1\right|\ge0\) Để \(\left|2x-1\right|=x+4\Leftrightarrow x+4\ge0\Rightarrow x\ge-4\)

Ta có : \(\left|2x-1\right|=\hept{\begin{cases}2x-1\Leftrightarrow2x-1\ge0\Rightarrow x\ge\frac{1}{2}\\1-2x\Leftrightarrow2x-1< 0\Rightarrow x< \frac{1}{2}\end{cases}}\)

TH1 : Với x ≥ 1/2 thì 2x - 1 = x + 4 <=> 2x - x = 4 + 1 => x = 5 ( TM )

TH2 : Với x < 1/2 thì 1 - 2x = x + 4 <=> - 2x - x = 4 - 1 <=> - 3x = 3 => x = - 1 ( TM )

vẬY X = { - 1; 5 }

|2x-1|=x+4

<=>2x-1=x+4

=>2x-x=4+1

=>x=5