Giải:

a) \(\dfrac{-1}{5}\le\dfrac{x}{8}\le\dfrac{1}{4}\) 

\(\Rightarrow\dfrac{-8}{40}\le\dfrac{5x}{40}\le\dfrac{10}{40}\) 

\(\Rightarrow5x\in\left\{0;\pm5;10\right\}\) 

\(\Rightarrow x\in\left\{0;\pm1;2\right\}\) 

b) \(\dfrac{4}{x-6}=\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow\dfrac{4}{x-6}=\dfrac{-12}{18}\) 

\(\Rightarrow-12.\left(x-6\right)=4.18\) 

\(\Rightarrow-12x+72=72\) 

\(\Rightarrow-12x=72-72\) 

\(\Rightarrow-12x=0\) 

\(\Rightarrow x=0:-12\) 

\(\Rightarrow x=0\) 

\(\Rightarrow\dfrac{y}{24}=\dfrac{-12}{18}\) 

\(\Rightarrow y=\dfrac{-12.24}{18}=-16\) 

c) \(\dfrac{x+46}{20}=x.\dfrac{2}{5}\) 

\(\dfrac{x+46}{20}=\dfrac{2x}{5}\) 

\(\Rightarrow5.\left(x+46\right)=2x.20\) 

\(\Rightarrow5x+230=40x\) 

\(\Rightarrow5x-40x=-230\) 

\(\Rightarrow-35x=-230\) 

\(\Rightarrow x=-230:-35\) 

\(\Rightarrow x=\dfrac{46}{7}\) 

Chúc bạn học tốt!

Bài 13: 

a: =>20-x=15-8+13=20

hay x=0

❤                                              ✔

a) Tìm số nguyên x, biết:

Với \(x\in Z\), ta có:

9 - x = 8 - (2x + 16)

<=> 9 - x = 8 - 2x - 16 

<=> 9 - x = -2x - 8

<=> x  = -17 (TM)

Vậy x = -17

b) Với \(x\in Z\), ta có:

18 - 2x = 21 - (3x - 5)

<=> 18 - 2x = 21 - 3x + 5

<=> 18 - 2x = 26 - 3x

<=> x = 8 (TM)

Vậy x = 8

a) Ta có: 9-x=8-(2x+16)

\(\Leftrightarrow9-x=8-2x-16\)

\(\Leftrightarrow9-x=-2x-8\)

\(\Leftrightarrow9-x+2x+8=0\)

\(\Leftrightarrow x+17=0\)

hay x=-17

Vậy: x=-17

b) Ta có: 18-2x=21-(3x-5)

\(\Leftrightarrow18-2x=21-3x+5\)

\(\Leftrightarrow18-2x+3x-26=0\)

\(\Leftrightarrow x-8=0\)

hay x=8

Vậy: x=8

a,\(\dfrac{6}{2x+1}=\dfrac{2}{7}\)

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

b: \(\Leftrightarrow x+8\in\left\{1;-1;5;-5\right\}\)

hay \(x\in\left\{-7;-9;-3;-13\right\}\)

a) 

\(\left(x+1\right)\left(y-2\right)=5\\ \Rightarrow\left(x+1\right),\left(y-2\right)\inƯ\left(5\right)=\left\{1;-1;5;-5\right\}\)

Ta có bảng:

Vậy \(\left(x;y\right)=\left(0;7\right),\left(-2;-3\right),\left(4;3\right),\left(-6;1\right)\)

b) 

\(\left(x-5\right)\left(y+4\right)=-7\\ \Rightarrow\left(x-5\right),\left(y+4\right)\inƯ\left(-7\right)=\left\{1;-1;7;-7\right\}\)

Ta có bảng:

Vậy \(\left(x;y\right)=\left(6;-11\right),\left(4;3\right),\left(12;-5\right),\left(-2;-3\right)\)

e) 

\(x-\left(17-8\right)=5+\left(10-3x\right)\\ \Rightarrow x-9=5+10-3x\\ \Rightarrow x+3x=5+10+9\\ \Rightarrow4x=24\\ \Rightarrow x=\dfrac{24}{4}=6\)

Vậy \(x=6\)

giúp mình với nhé,mình đang cần gấp

\(a,\left(8+x\right)\left(6-x\right)=0\\ \Rightarrow\left[{}\begin{matrix}8+x=0\\6-x=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-8\\x=6\end{matrix}\right.\\ b,x^2-5x=0\\ \Rightarrow x\left(x-5\right)=0\\ \Rightarrow\left[{}\begin{matrix}x=0\\x-5=0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0\\x=5\end{matrix}\right.\)

a) (8+x).(6-x)=0
<=> 8+x = 0 hoặc 6-x = 0
=> x = -8 hoặc x = 6

b) c) x^2 - 5x=0
<=> x^2 = 0 hoặc -5x = 0
=> x = 0 hoặc x = 5

a) [8+x]x[6-x]=0

vì tích của chúng phải bằng 0 nên có 2 trường hợp

TH1:(8+(-8))x(6-(xEN))

TH2:(8+(XEN))x(6-6)

`a)`

`x-15 = -21`

`<=> x = -21 + 15`

`<=> x = -6`

`b)`

`42 - x = -7`

`<=> x = 42 - (-7)`

`<=> x = 42 + 7`

`<=> x = 49`

`c)`

`12 - (30-x) = -23`

`30-x = 12 - (-23)`

`30-x = 35`

`x = 30-35`

`x = -5`

`d)`

`31 - (17+x)=18`

`<=> 17+x = 31-18`

`<=> 17+x =13`

`<=> x = 13-17`

`<=> x = -4`

a) \(x-15=-21\)

\(\Rightarrow x=-21+15=-6\)

b) \(42-x=-7\)

\(\Rightarrow x=42+7=49\)

c) \(12-\left(30-x\right)=-23\)

\(\Rightarrow30-x=12+23=35\)

\(\Rightarrow x=30+35=65\)

d) \(31-\left(17+x\right)=18\)

\(\Rightarrow17+x=31-18=13\)

\(\Rightarrow x=13-17=-4\)

a) $x-15=-21$

$\Rightarrow x=-21+15=-6$

b) $42-x=-7$

$x=42+7=49$

c) $12-\left(30-x\right)=-23$

$30-x=12+23=35$

$x=30+35=65$

d) $31-\left(17+x\right)=18$

$17+x=31-18=13$

$x=13-17=-4$