Bài 2 : a, x = -36/9 = -4

b, đề sai 

c, <=> -2 =< x =< -3 => x = -1 

Bài 1: 

a: 2/8=9/36; 2/9=8/36; 8/2=36/9; 9/2=36/8

b: -2/4=9/-18; -2/9=4/-18; 4/-2=-18/9; 9/-2=-18/4

Bài 2: 

a: =>x/3=-4/3

hay x=-4

Câu b đề sai rồi bạn

giúp mình với ạ mình cần gấp

a) Ta có: \(\dfrac{x}{2}=\dfrac{y}{5}\)

⇒\(\dfrac{y-x}{5-2}=\dfrac{6}{3}=2\)

\(\dfrac{x}{2}=2\Rightarrow x=4\)

\(\dfrac{y}{5}=2\Rightarrow y=10\)

\(\dfrac{z}{10}=2\Rightarrow z=20\)

b) Ta có: \(\dfrac{x}{8}=\dfrac{2y}{6}=\dfrac{z}{7}\)

\(\dfrac{x-2y+z}{8-6+7}=\dfrac{18}{9}=2\)

\(\dfrac{x}{8}=2\Rightarrow x=16\)

\(\dfrac{y}{3}=2\Rightarrow y=6\)

\(\dfrac{z}{7}=2\Rightarrow z=14\)

-29-9(2x-1)\(^2\)= -110

(=) 9(2x-1)² = (-29) +110

(=) 9(2x-1)² = 81

(=) (2x-1)² =81: 9

(=) (2x-1)² =9

(=) (2x-1)² = 3² =(-3)²

\(\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}}\)

\(\orbr{\begin{cases}2x=4\\2x=-2\end{cases}}\)

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

vậy : ........

a,\(-29-9\left(2x-1\right)^2=-110\)

\(=>-29+110=9.\left(2x-1\right)^2\)

\(=>81=9.\left(2x-1\right)^2\)

\(=>\left(2x-1\right)^2=9\)

\(=>\orbr{\begin{cases}2x-1=3\\2x-1=-3\end{cases}=>\orbr{\begin{cases}x=\frac{4}{2}=2\\x=\frac{-2}{2}=-1\end{cases}}}\)

b,\(-2.\left(-x-5\right)+28=20-3.4\)

\(=>2x+10+28=20-12=8\)

\(=>2x=8-28-10=-30\)

\(=>x=\frac{-30}{2}=-15\)

Bài 4: 

b: Ta có: \(2x\left(x-\dfrac{1}{4}\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x=\dfrac{1}{4}\end{matrix}\right.\)

a: \(\Leftrightarrow\dfrac{x}{-4}=\dfrac{21}{y}=\dfrac{z}{-80}=\dfrac{3}{4}\)

=>x=-3; y=28; z=-60

b: 5/12=x/-72

=>x=-72*5/12=-6*5=-30

c: =>x+3=-5

=>x=-8

a) \(\dfrac{5}{x}=\dfrac{-10}{12}.\Rightarrow x=-6.\)

b) \(\dfrac{4}{-6}=\dfrac{x+3}{9}.\Rightarrow x+3=-6.\Leftrightarrow x=-9.\)

c) \(\dfrac{x-1}{25}=\dfrac{4}{x-1}.\left(đk:x\ne1\right).\Leftrightarrow\dfrac{x-1}{25}-\dfrac{4}{x-1}=0.\)

\(\Leftrightarrow\dfrac{x^2-2x+1-100}{25\left(x-1\right)}=0.\Leftrightarrow x^2-2x-99=0.\)

\(\Leftrightarrow\left[{}\begin{matrix}x=11.\\x=-9.\end{matrix}\right.\) \(\left(TM\right).\)

a) \(x-\dfrac{2}{3}=\dfrac{3}{8}\Rightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{25}{24}\)

b) \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\Rightarrow x-\dfrac{3}{4}=\dfrac{1}{4}\Rightarrow x=\dfrac{1}{4}+\dfrac{3}{4}=1\)

c) \(\dfrac{3}{2}-\left(x+\dfrac{1}{2}\right)=\dfrac{4}{5}\Rightarrow x+\dfrac{1}{2}=\dfrac{3}{2}-\dfrac{4}{5}=\dfrac{7}{10}\)

\(\Rightarrow x=\dfrac{7}{10}-\dfrac{1}{2}=\dfrac{1}{5}\)

d) \(\left|x-2\right|-1=0\Rightarrow\left|x-2\right|=1\)

\(\Rightarrow\left[{}\begin{matrix}x-2=1\\x-2=-1\end{matrix}\right.\)\(\Rightarrow\left[{}\begin{matrix}x=3\\x=1\end{matrix}\right.\)

a: Ta có: \(x-\dfrac{2}{3}=\dfrac{3}{8}\)

\(\Leftrightarrow x=\dfrac{3}{8}+\dfrac{2}{3}=\dfrac{9}{24}+\dfrac{16}{24}=\dfrac{25}{24}\)

b: Ta có: \(x-\dfrac{3}{4}=\dfrac{13}{10}:\dfrac{26}{5}\)

\(\Leftrightarrow x-\dfrac{3}{4}=\dfrac{13}{10}\cdot\dfrac{5}{26}=\dfrac{1}{4}\)

hay x=1