Bài 1: Tính hợp lí (nếu có thể) Bài 2:Tìm x biết
a, \(\left(4\frac{2}{3}-1\frac{1}{6}\right):\left(1,75+1\frac{1}{2}\right)\) a, \(\frac{4}{9}+x=\frac{-5}{3}\)
b, \(\frac{11}{53}+\left(\frac{32}{47}+\frac{-10}{53}\right)+\frac{-64}{94}+\frac{1}{-53}+\frac{1}{3}\) b, 2,4:\(\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)
c, 125% -7\(\frac{1}{2}+0,5:\frac{4}{3}\) c, \(\frac{x+1}{-8}=\frac{-2}{x+1}\)
d, 1\(\frac{1}{20}.\left(\frac{-2}{3}\right)^2+\left(0,8-\frac{8}{15}\right):\frac{-4}{17}\) d, \(\left(\frac{2}{1.3}+\frac{2}{3.5}+\frac{2}{5.7}+...+\frac{2}{97.99}\right)-x=\frac{-100}{99}\)
Bài 2:
a) \(\frac{4}{9}+x=\frac{-5}{3}\)
\(\Leftrightarrow x=\frac{-5}{3}-\frac{4}{9}\)
\(\Leftrightarrow x=\frac{-15}{9}-\frac{4}{9}\)\(=\frac{-19}{9}\)
Vậy: \(x=\frac{-19}{9}\)
b) \(2,4:\left(\frac{1}{2}.x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{24}{10}:\left(\frac{1}{2}x-\frac{3}{4}\right)=\frac{3}{10}\)
\(\Leftrightarrow\frac{1}{2}x-\frac{3}{4}=\frac{24}{10}:\frac{3}{10}=\frac{24}{10}.\frac{10}{3}\)\(=8\)
\(\Leftrightarrow\frac{1}{2}x=8+\frac{3}{4}=\frac{35}{4}\)
\(\Leftrightarrow x=\frac{35}{4}:\frac{1}{2}=\frac{35}{4}.2=\frac{35}{2}\)
c) \(\frac{x+1}{-8}=\frac{-2}{x+1}\)
\(\Rightarrow\left(x+1\right).\left(x+1\right)=\left(-2\right).\left(-8\right)\)
\(\Leftrightarrow\left(x+1\right)^2=16=4^2=\left(-4\right)^2\)
\(\Rightarrow\left[{}\begin{matrix}x+1=4\\x+1=-4\end{matrix}\right.\)
\(\Leftrightarrow\left[{}\begin{matrix}x=3\\x=-5\end{matrix}\right.\)
Vậy: \(x\in\left\{3;-5\right\}\)
