a: \(=\dfrac{1.3-1.6}{2.6}-\dfrac{5}{6}:2\)

\(=-\dfrac{3}{26}-\dfrac{5}{12}\)

\(=\dfrac{-83}{156}\)

b: \(=20.8\left(-15.5+3.5\right)+9.2\left(3.5-15.5\right)\)

\(=-12\left(20.8+9.2\right)\)

\(=-12\cdot30=-360\)

\(A=\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{-\frac{131}{100}}{\frac{13}{5}}-\frac{5}{6}:2\)

\(=-\frac{131}{260}-\frac{5}{12}=-\frac{359}{390}\)

\(B=\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}.\frac{26}{75}=\frac{13}{12}\)

Ta có : \(A=-\frac{359}{390}\approx-0,9\)

\(B=\frac{13}{12}\approx1,08\)

\(\Rightarrow A< x< B\) mà x nguyên \(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

Ta có:

\(A=\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{\frac{-13}{10}}{\frac{13}{5}}-\frac{5}{6}:2=\frac{-1}{2}-\frac{5}{12}=\frac{-11}{12}\)

\(B=\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}:\frac{75}{26}=\frac{13}{12}\)

\(\Rightarrow A< x< B\Rightarrow\frac{-11}{12}< x< \frac{13}{12}\Rightarrow-1< x\le1\Rightarrow x\in\left\{0;1\right\}\)

A= \(\frac{1,11+0,19-1,3.2}{2,06+0,54}-\left(\frac{1}{2}+\frac{1}{3}\right):2=\frac{-\frac{131}{100}}{\frac{13}{5}}-\frac{5}{6}:2\)

=\(-\frac{131}{260}-\frac{5}{12}=-\frac{359}{390}\)

B= \(\left(5\frac{7}{8}-2\frac{1}{4}-0,5\right):2\frac{23}{26}=\left(\frac{47}{8}-\frac{9}{4}-\frac{1}{2}\right):\frac{75}{26}=\frac{25}{8}.\frac{26}{75}=\frac{13}{12}\)

b) ta có : A=\(-\frac{359}{390}\approx-0,9\)

B= \(\frac{13}{12}\approx1,08\)

=> A<x<B mà x nguyên => x=0 hoặc x=1

e) \(\frac{1}{100}-\frac{1}{100.99}-\frac{1}{99.98}-...-\frac{1}{3.2}-\frac{1}{2.1}\)

\(=\frac{1}{100}-\left(\frac{1}{100.99}+\frac{1}{99.98}+...+\frac{1}{3.2}+\frac{1}{2.1}\right)\)

\(=\frac{1}{100}-\left(\frac{1}{1}-\frac{1}{2}+\frac{1}{2}-\frac{1}{3}+...+\frac{1}{98}-\frac{1}{99}+\frac{1}{99}-\frac{1}{100}\right)\)

\(=\frac{1}{100}-\left(1-\frac{1}{100}\right)\)

\(=\frac{1}{100}-\frac{99}{100}\)

\(=\frac{-98}{100}\)\

\(=\frac{-49}{50}\)

g)-15,5 . 20,8 + 3,5.9,2 - 15,5.9,2 + 3,5.20,8

=20,8.(-15.5+3,5)+9,2(-15.5+3.5)

=(-15.5+3.5)(20.8+9.2)

=(-12).30

=-360

a) -49/50

b ) -360

hok tốt

a: \(A=\dfrac{1.3-26}{2.6}-\dfrac{3}{8}\cdot\dfrac{1}{2}=-\dfrac{19}{2}-\dfrac{3}{16}=-\dfrac{155}{16}\)

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

\(=\dfrac{47-18-4}{8}\cdot\dfrac{26}{75}=\dfrac{1}{3}\cdot\dfrac{13}{4}=\dfrac{13}{12}\)

c: Để A<x<B thì \(-\dfrac{155}{16}< x< \dfrac{13}{12}\)

=>-10<x<2

hay \(x\in\left\{-9;-8;-7;...;0;1\right\}\)