a)\(\frac{-5}{3}< x< \frac{-24}{35}.-\frac{5}{6}\)
\(\Rightarrow\frac{-5}{3}< x< \frac{4}{7}\)
\(\Rightarrow\frac{-35}{21}< x< \frac{12}{21}\)
\(\Rightarrow x\in\left\{\frac{-34}{21};\frac{-33}{21};\frac{-32}{21};...;\frac{10}{21};\frac{11}{21};\frac{12}{21}\right\}\)
Mà \(x\inℤ\) nên \(x\in\left\{-1;0;1\right\}\)
\(a,\left[\frac{-5}{3}\right]< x< \frac{-24}{35}\cdot\left[\frac{-5}{6}\right]\)
\(\Rightarrow\frac{-5}{3}< x< \frac{-24\cdot(-5)}{35\cdot6}\)
\(\Rightarrow\frac{-5}{3}< x< \frac{-4\cdot(-1)}{7\cdot1}=\frac{4}{7}\)
\(\Rightarrow\frac{5}{3}< x< \frac{4}{7}\)
\(\Rightarrow\frac{35}{21}< x< \frac{12}{21}\)
Mà \(x\inℤ\)nên \(x\in\varnothing\)
Vậy không có giá trị x thỏa mãn
\(b,(7x-11)^3=(-3)^2\cdot15+208\)
\(\Rightarrow(7x-11)^3=9\cdot15+208\)
\(\Rightarrow(7x-11)^3=343\)
\(\Rightarrow(7x-11)^3=7^3\)
\(\Rightarrow7x-11=7\)
\(\Rightarrow7x=7+11\)
\(\Rightarrow7x=18\)
\(\Rightarrow x=\frac{18}{7}\)
=> x không có giá trị thỏa mãn vì \(x\inℤ\)
Lộn phần a dòng thứ 3,4,...
Mk sửa lại :
\(\frac{-5}{3}< x< \frac{4}{7}\)
\(\Rightarrow\frac{-35}{21}< x< \frac{12}{21}\)
\(\Rightarrow-35< x< 12\)
\(\Rightarrow x\in\left\{-34;-33;...;11\right\}\)