vào thống kê xem link nhé: 

Câu hỏi của Kim Trân Ni - Toán lớp 8 - Học toán với OnlineMath

a/ \(\frac{2}{3}+\frac{4}{35}< \frac{x}{105}< \frac{1}{7}+\frac{2}{5}+\frac{1}{3}\)

\(\Rightarrow\frac{82}{105}< \frac{x}{105}< \frac{92}{105}\)

\(\Rightarrow82< x< 92\)

\(\Rightarrow x=\left\{83;84;85;86;87;88;89;90;91\right\}\)

b/ \(-\frac{7}{15}+\frac{8}{60}+\frac{24}{90}\le\frac{x}{15}\le\frac{3}{5}+\frac{8}{30}+-\frac{4}{10}\)

\(\Rightarrow-\frac{1}{15}\le\frac{x}{15}\le\frac{7}{15}\)

\(\Rightarrow-1\le x\le7\)

\(\Rightarrow x=\left\{-1;0;1;2;3;4;5;6;7\right\}\)

Ta có : \(-\frac{5}{6}+\frac{8}{3}+\frac{29}{-6}=-3\) và \(\frac{1}{2}+2+\frac{5}{2}=5\)

Vậy -3 < x < 5. Do x \(\in\) Z nên x \(\in\) {-2; -1; 0; 1; 2; 3; 4}

Ta có :\(\frac{x-1}{-2}=\frac{-8}{x-1}\Rightarrow\left(x-1\right)^2=16\Rightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}}\)

\(\frac{x-1}{-2}=\frac{-8}{x-1}\left(x\ne1\right)\)

\(\Leftrightarrow\left(x-1\right)\left(x-1\right)=-2\cdot\left(-8\right)\)

\(\Leftrightarrow\left(x-1\right)^2=16\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}\Leftrightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}\left(tm\right)}}\)

Vậy x=5; x=-3

\(\frac{x-1}{-2}=\frac{-8}{x-1}\)

\(\Rightarrow\left(x-1\right)\left(x-1\right)=\left(-2\right)\cdot\left(-8\right)\)

\(\Rightarrow\left(x-1\right)^2=16\)

\(\Rightarrow\orbr{\begin{cases}\left(x-1\right)^2=4^2\\\left(x-1^2\right)=\left(-4\right)^2\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x-1=4\\x-1=-4\end{cases}}\)

\(\Rightarrow\orbr{\begin{cases}x=5\\x=-3\end{cases}}\)