Question

[Thtamm]

 làm 4A,4B ạ

Answer 1

[Toru]

4A:

\(\left[\left(\dfrac{2}{193}-\dfrac{3}{386}\right).\dfrac{193}{17}+\dfrac{33}{34}\right]:\left[\left(\dfrac{7}{1931}+\dfrac{11}{3862}\right).\dfrac{1931}{25}+\dfrac{9}{2}\right]\)

\(=\left(\dfrac{2}{193}.\dfrac{193}{17}-\dfrac{3}{386}.\dfrac{193}{17}+\dfrac{33}{34}\right):\left(\dfrac{7}{1931}.\dfrac{1931}{25}+\dfrac{11}{3862}.\dfrac{1931}{25}+\dfrac{9}{2}\right)\)

\(=\left(\dfrac{2}{17}-\dfrac{3}{34}+\dfrac{33}{34}\right):\left(\dfrac{7}{25}+\dfrac{11}{50}+\dfrac{9}{2}\right)\)

\(=\left(\dfrac{4}{34}+\dfrac{30}{34}\right):\left(\dfrac{14}{50}+\dfrac{11}{50}+\dfrac{9}{2}\right)\)

\(=\dfrac{34}{34}:\left(\dfrac{1}{2}+\dfrac{9}{2}\right)\)

\(=1:\dfrac{10}{2}=\dfrac{1}{5}\)

4B:

a, \(A=\dfrac{1,11+0,19-13.2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\)

\(=\dfrac{1,3-26}{2,6}-\dfrac{3}{4}:2\)

\(=\dfrac{-24,7}{2,6}-\dfrac{3}{8}=\dfrac{-19}{2}-\dfrac{3}{8}=-\dfrac{79}{8}\)

\(B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):2\dfrac{23}{26}\)

\(=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)

\(=\left(\dfrac{29}{8}-\dfrac{1}{2}\right).\dfrac{26}{75}\)

\(=\dfrac{25}{8}.\dfrac{26}{75}=\dfrac{13}{12}\)

b, Để \(A< x< B\) thì \(-\dfrac{79}{8}< x< \dfrac{13}{12}\)

\(\Rightarrow x\in\left\{-9;-8;-7;...;1\right\}\) (vì \(x\in\mathbb{Z}\))

$Toru$