a)\(\dfrac{4}{x}=\dfrac{x}{16}\)

<=>\(x^2=4.16=64\)

<=>\(x=\pm8\)

<=>x=-8(vì x<0)

b)\(\dfrac{x}{-24}=\dfrac{-6}{x}\)

<=>\(x^2=\left(-24\right)\left(-6\right)=144\)

<=>\(x=\pm12\)

<=>x=12(Vì x>0)

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!

a)\(x-5=-1\)

⇔\(x=4\)

b)\(x+30=-4\)

⇔\(x=-34\)

c)\(x-\left(-24\right)=3\)

⇔\(x+24=3\)

⇔\(x=-21\)

e)\(\left(x+5\right)+\left(x-9\right)=x+2\)

⇔\(x+5+x-9-x-2=0\)

⇔\(x-6=0\)

⇔\(x=6\)

f)\(\left(27-x\right)+\left(15+x\right)=x-24\)

⇔\(27-x+15+x-x+24=0\)

⇔\(66-x=0\)

⇔\(x=66\)

\(a.x-5=-1\)                      \(b.x+30=-4\)

\(x=\left(-1\right)+5\)                        \(x=\left(-4\right)-30\)

\(x=4\)                                    \(x=-34\)

\(c.x-\left(-24\right)=3\)                    \(e.\left(x+5\right)+\left(x-9\right)=x+2\)

\(x=3+\left(-24\right)\)                         \(x+5+x-9=x+2\)

\(x=-21\)                                  \(2x-4=x+2\)

                                                \(2x-x=2+4\)

                                                 \(x=6\)

\(f.\left(27-x\right)+\left(15+x\right)=x-24\)

\(27-x+15+x=x-24\)

\(27+15=x-24\)

\(42=x-24\)

\(x=24+42\)

\(x=66\)

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

⇒\(\dfrac{6}{2x+1}=\dfrac{6}{21}\)

⇒\(2x+1=21\)

\(2x=21-1\)

\(2x=20\)

⇒\(x=10\)

x=\(-\frac{125}{27}\)

\(x=-\frac{125}{27}\)

x = âm 125 / 27

k đii

a,

7x – 8 = 713

7x = 713 + 8

7x = 721

x = 721 : 7

x = 103.

b,

2448:[119-(x-6)]=24

119-(x-6)= 2448 : 24

119-(x-6)= 102

x-6= 119-102

x-6= 17

x=17+6

x=23

Bài 10:

a: 2x-3 là bội của x+1

=>\(2x-3⋮x+1\)

=>\(2x+2-5⋮x+1\)

=>\(-5⋮x+1\)

=>\(x+1\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{0;-2;4;-6\right\}\)

b: x-2 là ước của 3x-2

=>\(3x-2⋮x-2\)

=>\(3x-6+4⋮x-2\)

=>\(4⋮x-2\)

=>\(x-2\inƯ\left(4\right)\)

=>\(x-2\in\left\{1;-1;2;-2;4;-4\right\}\)

=>\(x\in\left\{3;1;4;0;6;-2\right\}\)

Bài 14:

a: \(4n-5⋮2n-1\)

=>\(4n-2-3⋮2n-1\)

=>\(-3⋮2n-1\)

=>\(2n-1\inƯ\left(-3\right)\)

=>\(2n-1\in\left\{1;-1;3;-3\right\}\)

=>\(2n\in\left\{2;0;4;-2\right\}\)

=>\(n\in\left\{1;0;2;-1\right\}\)

mà n>=0

nên \(n\in\left\{1;0;2\right\}\)

b: \(n^2+3n+1⋮n+1\)

=>\(n^2+n+2n+2-1⋮n+1\)

=>\(n\left(n+1\right)+2\left(n+1\right)-1⋮n+1\)

=>\(-1⋮n+1\)

=>\(n+1\in\left\{1;-1\right\}\)

=>\(n\in\left\{0;-2\right\}\)

mà n là số tự nhiên

nên n=0

thiếu bài 16