3x – 47.2 + x = - 29 + (-9)
3x – 47.2 + x = -38
3x – 94 +    x = -38
3x – 94   = -38 - x
3x + x     = -38 + 94
        4x   = 56
          x   = 56 : 4
          x   = 14
Vậy x = 14

\(3x-47.2+x=-29+\left(-9\right)\\ \Rightarrow4x-94=-38\\ \Rightarrow4x=-38+94\\ \Rightarrow4x=56\\ \Rightarrow x=56:4\\ \Rightarrow x=14\)

Tìm số nguyên biết:

A) 2x-35=15

2x=15+35

2x=50

x=50:2

x=25.

sao không làm hết!

À vì hôm qua mình phải làm Violympic nên không có thời gian ấy mà.

a) Vì 29 là số nguyên tố nên :

(x-9).(y-18)=1.29=29.1 

+)Nếu (x-9)(y-18)=1.29

\(\hept{\begin{cases}\\\end{cases}}\begin{cases}x-9=1\\y-18=29\end{cases}\)

\(\Leftrightarrow\hept{\begin{cases}\\\end{cases}}\begin{cases}x=10\\y=47\end{cases}\)

+)Nếu (x-9)(y-18)=29.1

\(\hept{\begin{cases}\\\end{cases}}\begin{cases}x-9=29\\y-18=1\end{cases}\)

\(\Rightarrow\hept{\begin{cases}\\\end{cases}}\begin{cases}x=38\\y=19\end{cases}\)

Vậy...

Câu b làm tương tự nhé bạn

|3x - 2| + 5 = 9 - x

|3x - 2| = 4 - x

3x - 2 = 4 - x hoặc 3x - 2 = x - 4

3x + x = 4 + 2 hoặc 3x - x = - 4 + 2

4x = 6 hoặc 2x = - 2

x = 1,5 hoặc x = - 1

mà x \(\in\) Z nên x = - 1

a) x = 4

b) x = -8

c) | x | - 9 = -2 + 17

| x | = 15 + 9

| x | = 24

x = 24 hoặc x = -24

d) |x – 9| = -2 + 17

|x – 9| = 15

x – 9 = 15 hoặc x – 9 = -15

x = 24 hoặc x = -6

\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow x\left(\frac{1}{2}-\frac{2}{3}\right)=\frac{7}{12}\Rightarrow x\cdot\left(\frac{3}{6}-\frac{4}{6}\right)=\frac{7}{12}\)

\(\Rightarrow x\cdot\left(-1\right)=\frac{7}{12}\Rightarrow x=\frac{7}{12}:\left(-1\right)=\frac{7}{-12}\)

\(c,\frac{\left(x-5\right)}{12}\cdot\frac{9}{29}=\frac{-6}{29}\Rightarrow\frac{\left(x-5\right)}{12}=\frac{-6}{29}:\frac{9}{26}\)

\(\frac{\Rightarrow\left(x-5\right)}{12}=\frac{-6}{9}=\frac{-2}{3}\Rightarrow x-5=-\frac{2}{3}\cdot12\)

\(\Rightarrow x-5=\frac{-24}{3}=-8\Rightarrow x=-8+5=-3\)

\(a,0,5x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow\frac{1}{2}x-\frac{2}{3}x=\frac{7}{12}\)

\(\Rightarrow-\frac{1}{6}x=\frac{7}{12}\)

\(\Rightarrow x=-\frac{7}{2}\)

\(c,\frac{x-5}{12}\cdot\frac{9}{29}=-\frac{6}{29}\)

\(\Rightarrow\frac{x-5}{12}=-\frac{2}{3}\)

\(\Rightarrow x-5=12.\left(-\frac{2}{3}\right)\)

\(\Rightarrow x-5=-8\)

\(\Rightarrow x=-3\)

b)\(\frac{-8}{17}+\frac{5}{17}< \frac{x}{17}< \frac{-6}{17}+\frac{9}{17}\)\(\left(x\in Z\right)\)

\(\Rightarrow\frac{-3}{17}< \frac{x}{17}< \frac{3}{17}\)

\(\Rightarrow-3< x< 3\)

\(\Rightarrow x\in\left\{-2;-1;0;1;2\right\}\)

Vậy \(x\in\left\{-2;-1;0;1;2\right\}\)

`\color{grey}\text{#071931}`

`(4 + 2x) * (9 - 3x) = 0`

TH1: `4 + 2x = 0 => 2x = -4 => x = -2`

TH2: `9 - 3x = 0 => 3x = 9 => x = 3`

Vậy, `x \in {-2; 3}`

____

`(3x - 19)^3 = 125`

`=> (3x - 19)^3 = 5^3`

`=> 3x - 19 = 5`

`=> 3x = 24`

`=> x = 8`

Vậy, `x = 8.`

\(a,3x-75=45+3.\left(-25\right)\)

\(3x-75=-30\)

\(3x=45\)

\(x=15\)

\(b,2.\left(-7\right)-11=6x+17\)

\(-14-11-17=6x\)

\(-42=6x\)

\(-7=x\Leftrightarrow x=-7\)

\(d,-4.|x-3|=-48\)

\(|x-3|=12\)

\(\Rightarrow\orbr{\begin{cases}x-3=12\\x-3=-12\end{cases}\Rightarrow\orbr{\begin{cases}x=15\\x=-9\end{cases}}}\)