Bài 1:

\(a)\left(\dfrac{-28}{29}\right).\left(\dfrac{-38}{16}\right)=\dfrac{\left(-28\right).\left(-38\right)}{29.16}=\dfrac{1064}{464}=\dfrac{133}{58}\)

\(b)\left(\dfrac{-21}{16}\right).\left(\dfrac{-24}{7}\right)=\dfrac{\left(-21\right).\left(-24\right)}{16.7}=\dfrac{504}{112}=\dfrac{9}{2}\)

\(c)\left|\dfrac{-12}{17}\right|.\left(\dfrac{-34}{9}\right)=\dfrac{12}{17}.\left(\dfrac{-34}{9}\right)=\dfrac{12.\left(-34\right)}{17.9}=\dfrac{-408}{153}=\dfrac{-8}{3}\)

Bài 3:

\(a)\left|x\right|=21\)

\(\Rightarrow\left[{}\begin{matrix}x=-21\\x=21\end{matrix}\right.\)

\(b)\left|x\right|=\dfrac{17}{9};x< 0\)

\(\Rightarrow x=\dfrac{-17}{9}\)

\(c)\left|x\right|=1\)

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

\(\left|x\right|=\dfrac{2}{5}\)

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

\(d)\left|x\right|=0,35;x>0\)

\(\Rightarrow x=0,35\)

Bài 4:

\(a)\left|x\right|-1,7=2,3\)

\(\Rightarrow\left[{}\begin{matrix}x-1,7=2,3\\x-1,7=-2,3\end{matrix}\right.\)

\(\Rightarrow\left[{}\begin{matrix}x=4\\x=\dfrac{-3}{5}\end{matrix}\right.\)

\(b)\left|x\right|+\dfrac{3}{4}-\dfrac{1}{3}=0\)

\(\Rightarrow\left|x\right|+\dfrac{3}{4}=0+\dfrac{1}{3}\)

\(\Rightarrow\left|x\right|+\dfrac{3}{4}=\dfrac{1}{3}\)

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

\(\Rightarrow\left[{}\begin{matrix}x=\dfrac{-5}{12}\\x=\dfrac{-13}{12}\end{matrix}\right.\)

Chúc bạn học tốt!

a,/X/+/-5/=/-37/

X+5=37

      X=37-5

      X=32

\(\left(16-x\right)+\left|x+4\right|=0\)

Dấu ''='' xảy ra <=> \(\hept{\begin{cases}16-x=0\\x+4=0\end{cases}}\Leftrightarrow\hept{\begin{cases}x=16\\x=-4\end{cases}}\)

Vậy x = { 16 ; -4 }

Xl bạn mở ngoặc ra :   \(\left(x\in Z\right)\)ở dogf đầu tiền

và sử 2 dòng cuối đi nhé  \(\Rightarrow\orbr{\begin{cases}x=3\\x=-3\end{cases}}\)  ( thỏa mãn \(x\in Z\) )

Vậy x = \(\pm\) 3 

nhé... Xin lỗi bạn

- 14 . | x | + 2 = - 40

=> - 14 . | x | = - 42

=> | x | = 3

\(\Rightarrow\) \(x=\pm3\)

Vậy \(x=\pm3\)

Cảm ơn các bạn