a)x- [-2] = [-18]

\(x=\left(-18\right)+\left(-2\right)\)

\(x=-20\)

b) 2x- [+14]=[-14]

\(2x=\left(-14\right)+14\)

\(2x=0\)

\(x=0\)

c) [x+4] +5=20-(-12-7)

\(\left(x+4\right)+5=39\)

\(x+4=39-5\)

\(x+4=34\)

\(x=30\)

d)15-[2-x]=(-2)²

^{\(15-\left(2-x\right)=4\)}

^\(2-x=11\)

\(x=-9\)

e)[15-x] +[-25]=[-55]

\(15-x=\left(-55\right)-\left(-25\right)\)

\(15-x=-30\)

\(x=15--30\)

\(x=45\)

g)[17-(-4)] +[-24-(-5)]=[-x+3]

\(-x+3=21+\left(-19\right)\)

\(-x+3=2\)

\(x=1\)

chúc bạn học tốt

a,x-[-2]=[-18]              

x         =18+2

x         =20

Vậy x thuộc{20}

b,2x-[+14]=[-14]

  2x-14     =14

 2x            =14+14

   2x         =28

  x            =28:2

  x            =14

Vậy x thuộc{14}

c,[x+4]+5=20-(-12-7)

  [x+4]+5=20-(-19)

  [x+4]+5=20+19

 [x+4]+5=39

  [x+4]   =39-5

 [x+4]   =34

TH1:x+4=34

        x    =34-4

       x     =30

TH2:x+4=-34

       x     =-34-4

      x      =-38

vậy x thuộc{30;-38}

sorry bạn nha mk ko có tg nên bn làm nốt hộ mk nhá

a) | 2x - 1 | = 1- 3x

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

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

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

\(\orbr{\begin{cases}x=-2\\5x=0\end{cases}}\Rightarrow\orbr{\begin{cases}x=-2\\x=0\end{cases}}\)

b) | 1 - 2x | = x + 1 

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

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

\(\orbr{\begin{cases}-3x=0\\-x=-2\end{cases}}\Rightarrow\orbr{\begin{cases}x=0\\x=2\end{cases}}\)

tương tự

a, Áp dụng t/c dtsbn:

\(5x=7y\Rightarrow\dfrac{x}{7}=\dfrac{y}{5}=\dfrac{y-x}{5-7}=\dfrac{2}{-2}=-1\\ \Rightarrow\left\{{}\begin{matrix}x=-7\\y=-5\end{matrix}\right.\)

b, Áp dụng t/c dtsbn:

\(\dfrac{x}{y}=\dfrac{7}{2}\Rightarrow\dfrac{x}{7}=\dfrac{y}{2}=\dfrac{x+y}{7+2}=\dfrac{-27}{9}=-3\\ \Rightarrow\left\{{}\begin{matrix}x=-21\\y=-6\end{matrix}\right.\)

c, \(\dfrac{x}{32}=\dfrac{2}{x}\Rightarrow x^2=2\cdot32=64\Rightarrow\left[{}\begin{matrix}x=8\\x=-8\end{matrix}\right.\)

d, \(\left|x+\dfrac{1}{3}\right|-2=\dfrac{1}{2}\Rightarrow\left|x+\dfrac{1}{3}\right|=\dfrac{5}{2}\)

\(\Rightarrow\left[{}\begin{matrix}x+\dfrac{1}{3}=\dfrac{5}{2}\\x+\dfrac{1}{3}=-\dfrac{5}{2}\end{matrix}\right.\Rightarrow\left[{}\begin{matrix}x=\dfrac{13}{6}\\x=-\dfrac{17}{6}\end{matrix}\right.\)

\(a,\left|x+\dfrac{4}{5}\right|-\dfrac{1}{7}=0\\ \Rightarrow\left|x+\dfrac{4}{5}\right|=\dfrac{1}{7}\\ \Rightarrow\left[{}\begin{matrix}x+\dfrac{4}{5}=\dfrac{1}{7},\forall x+\dfrac{4}{5}\ge0\\x+\dfrac{4}{5}=-\dfrac{1}{7},\forall x+\dfrac{4}{5}< 0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35},\forall x\ge-\dfrac{4}{5}\left(N\right)\\x=-\dfrac{33}{35},\forall x< -\dfrac{4}{5}\left(N\right)\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=-\dfrac{23}{35}\\x=-\dfrac{33}{35}\end{matrix}\right.\)

\(b,\left|x-2\right|=x-2\\ \Rightarrow\left[{}\begin{matrix}x-2=x-2,\forall x-2\ge0\\x-2=2-x,\forall x-2< 0\end{matrix}\right.\\ \Rightarrow\left[{}\begin{matrix}x=0,\forall x\ge2\left(L\right)\\x=2,\forall x< 2\left(L\right)\end{matrix}\right.\\ \Rightarrow x\in\varnothing\)

bn viết tke bố tôi cx ko hiểu đc

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