a: \(x+\dfrac{3}{9}=\dfrac{7}{6}\cdot\dfrac{2}{3}\)

=>\(x+\dfrac{1}{3}=\dfrac{14}{18}=\dfrac{7}{9}\)

=>\(x=\dfrac{7}{9}-\dfrac{1}{3}=\dfrac{7}{9}-\dfrac{3}{9}=\dfrac{4}{9}\)

b: \(x-\dfrac{2}{3}=\dfrac{1}{8}:\dfrac{5}{4}\)

=>\(x-\dfrac{2}{3}=\dfrac{1}{8}\cdot\dfrac{4}{5}=\dfrac{1}{10}\)

=>\(x=\dfrac{1}{10}+\dfrac{2}{3}=\dfrac{3+20}{30}=\dfrac{23}{30}\)

TThế giới oi oi oi 

\(-x-2=\frac{5}{4}\)

\(\Rightarrow-x=\frac{5}{4}+2=\frac{13}{4}\)

\(\Rightarrow x=\frac{-13}{4}\)

\(-x-2=\frac{5}{4}\)

\(-x=\frac{5}{4}+2\)

\(-x=\frac{13}{4}\)

\(x=-\frac{13}{4}\)

-x - 2 = 5/4

=> -x = 5/4 + 2

=> -x = 13/4

=> x = -13/4

a. \(158-x=-12\)

\(x=158+12\)

\(x=170\)

b. \(37+x=12\)

\(x=12-37\)

\(x=-25\)

c. \(2x-15=-47\)

\(2x=\left(-47\right)+15\)

\(x=\dfrac{\left(-32\right)}{2}\)

\(x=-16\)

d. \(\left(-5\right)^2-\left(5x-3\right)=43\)

\(25-\left(5x-3\right)=43\)

\(\left(5x-3\right)=25-43\)

\(5x=\left(-18\right)+3\)

\(x=\dfrac{\left(-15\right)}{5}=-3\)

e. \(\left|x-1\right|+\left(-5\right)=2\)

\(\left|x-1\right|=2-\left(-5\right)\)

\(\left|x-1\right|=7\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\Leftrightarrow}\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right.\)

f. \(\left|x+1\right|=4\)

\(\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)

g. \(\left|x\right|=12\)

\(\Rightarrow x=\pm12\)

Câu h ko có x sao tìm

a) \(158-x=-12.\)

\(\Rightarrow x=158-\left(-12\right).\)

\(\Rightarrow x=158+12=170.\)

Vậy..........

b) \(37+x=12.\)

\(\Rightarrow x=12-37.\)

\(\Rightarrow x=-25.\)

Vậy..........

c) \(2x-15=-47.\)

\(\Rightarrow2x=-47+15.\)

\(\Rightarrow2x=-32.\)

\(\Rightarrow x=-\dfrac{32}{2}=-16.\)

Vậy..........

d) \(\left(-5\right)^2-\left(5x-3\right)=43.\)

\(\Rightarrow25-\left(5x-3\right)=43.\)

\(\Rightarrow5x-3=25-43.\)

\(\Rightarrow5x-3=-18.\)

\(\Rightarrow5x=-18+3.\)

\(\Rightarrow5x=-15.\)

\(\Rightarrow x=-\dfrac{15}{5}=3.\)

Vậy..........

e) \(\left|x-1\right|+\left(-5\right)=2.\)

\(\Rightarrow\left|x-1\right|=2-\left(-5\right).\)

\(\Rightarrow\left|x-1\right|=2+5.\)

\(\Rightarrow\left|x-1\right|=7.\)

\(\Leftrightarrow\left[{}\begin{matrix}x-1=7\\x-1=-7\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=8\\x=-6\end{matrix}\right..\)

Vậy..........

f) \(\left|x+1\right|=4\Leftrightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right..\)

Vậy..........

g) \(\left|x\right|=12\Leftrightarrow\left[{}\begin{matrix}x=12\\x=-12\end{matrix}\right..\)

Vậy..........

\(\left(x-1\right)\left(x+1\right)\left(x+2\right)=0\)

\(TH1:x-1=0\Leftrightarrow x=1\)

\(TH2:x+1=0\Leftrightarrow x=-1\)

\(TH3:x+2=0\Leftrightarrow x=-2\)

nhân đa thức vs đa thức , ko phải tìm x đâu bạn ạ! dù sao cững cảm ơn nh!

\(\dfrac{-10}{25}=\dfrac{-2}{5}\)

-1/25-9/25=-10/25=-2/5

\(=\dfrac{\left(-1\right)-9}{25}=\dfrac{\left(-1\right)+\left(-9\right)}{25}=\dfrac{-10}{25}=\dfrac{-2}{5}\)

Ta có:\(\frac{1}{6}x+\frac{1}{10}x-\frac{4}{15}x-1=0\)

     \(\Rightarrow\left(\frac{1}{6}+\frac{1}{10}-\frac{4}{15}\right)x=1\)

     \(\Rightarrow0x=1\)

      \(\Rightarrow x\in\varnothing\)

a) 4 ⋮ x ⇒ x ∈ Ư(4)

Ư(4) = {1;2;4}

⇒ x = {1;2;4}

b) -13 ⋮ (x+2) ⇒ x + 2 ∈ Ư(-13)

Ư(-13) = {1,-1,-13,13}

⇒ x = {-1,-3,-16;11}

Tìm x à bạn?

               x + 1 = ( x + 1 )² 

               x + 1 = x² + 2x + 1

               x - 2x - x² = - 1 + 1

               - x - x² = 0

                    - x ( x + 1) = 0

          TH1: - x = 0 suy ra x = 0

          TH2: x + 1 = 0 suy ra x = - 1

               Vậy x = 0 hoặc x = - 1.

x = 0 nha!

 chúc bn học tốt~

bạn đã kiểm tra kĩ chưa vậy?mình đọc đề câu B mà loạn não luôn á;-;

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