a.

\(A=\dfrac{1,11+0,19-13,2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\\ =\dfrac{2.2-13,2}{2,6}-\dfrac{3}{4}:2\\ =\dfrac{-11}{2,6}-\dfrac{3}{8}\\ =-\dfrac{55}{13}-\dfrac{3}{8}=-\dfrac{479}{104}\simeq-4,6\\ B=\left(5\dfrac{7}{8}-2\dfrac{1}{4}-0,5\right):2\dfrac{23}{26}\\ =\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):2\dfrac{23}{26}\\ =\dfrac{13}{12}=1.08\left(3\right)\)

a)
=\(2x-\dfrac{1}{2}=\dfrac{12}{9}\cdot\dfrac{3}{4}=1\)
=\(2x=1+\dfrac{1}{2}=1.5\)
=\(x=1.5:2=0.75\)
b)
=\(x^2=0+2=2\)
TH1:\(x=2\)
TH2:\(x=-2\)

Bài 1:

a: \(2x-\dfrac{1}{2}:\dfrac{3}{4}=\dfrac{12}{9}\)

=>\(2x-\dfrac{1}{2}\cdot\dfrac{4}{3}=\dfrac{4}{3}\)

=>\(2x=\dfrac{4}{3}+\dfrac{2}{3}=\dfrac{6}{3}=2\)

=>x=2/2=1

b: \(x^2-2=0\)

=>\(x^2=2\)

=>\(\left[{}\begin{matrix}x=\sqrt{2}\\x=-\sqrt{2}\end{matrix}\right.\)

Bài 2:

a: \(A=\dfrac{1,11+0,19-13\cdot2}{2,06+0,54}-\left(\dfrac{1}{2}+\dfrac{1}{4}\right):2\)

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

\(=-9,5-\dfrac{3}{8}=-\dfrac{79}{8}\)

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

\(=\left(5+\dfrac{7}{8}-2-\dfrac{1}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}\)

\(=\left(3+\dfrac{1}{8}\right)\cdot\dfrac{26}{75}=\dfrac{25}{8}\cdot\dfrac{26}{75}=\dfrac{13}{12}\)

b: A<x<B

=>\(-\dfrac{79}{8}< x< \dfrac{13}{12}\)

mà \(x\in Z\)

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

a: \(A=\dfrac{1.3-2.6}{2.6}-\dfrac{5}{6}\cdot\dfrac{1}{2}=\dfrac{-1}{2}-\dfrac{5}{12}=\dfrac{-11}{12}\)

\(B=\left(\dfrac{22}{3}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}=\dfrac{55}{12}\cdot\dfrac{26}{75}=\dfrac{143}{90}\)

b: Để A<x<B thì \(\dfrac{-11}{12}< x< \dfrac{143}{90}\)

mà x là số nguyên

nên \(x\in\left\{0;1\right\}\)

\(M=\left|x-2002\right|+\left|x-2001\right|\)\(=\left|x-2002\right|+\left|2001-x\right|\ge\left|x-2002+2001-x\right|=\left|-2002+2001\right|=1\)

tức \(M\ge1\) \(\Leftrightarrow\left[{}\begin{matrix}x-2001=0\\x-2002=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2001\\x=2002\end{matrix}\right.\)

Vậy Min_M = - 1 \(\Leftrightarrow\left[{}\begin{matrix}x=2001\\x=2002\end{matrix}\right.\)

a: \(A=\dfrac{1.3-2.6}{2.6}-\dfrac{5}{6}:2=\dfrac{-1}{2}-\dfrac{5}{12}=\dfrac{-11}{12}\)

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

\(=\dfrac{47-18-4}{8}\cdot\dfrac{26}{75}=\dfrac{25}{75}\cdot\dfrac{26}{8}=\dfrac{1}{3}\cdot\dfrac{13}{4}=\dfrac{13}{12}\)

b: Để A<x<B thì \(\dfrac{-11}{12}< x< \dfrac{13}{12}\)

mà x là số nguyên

nên \(x\in\left\{0;1\right\}\)

a: \(A=\dfrac{1.3-2.6}{2.6}-\dfrac{5}{6}:2=\dfrac{-1}{2}-\dfrac{5}{12}=\dfrac{-11}{12}\)

\(B=\left(\dfrac{47}{8}-\dfrac{9}{4}-\dfrac{1}{2}\right):\dfrac{75}{26}=\dfrac{47-18-4}{8}\cdot\dfrac{26}{75}=\dfrac{25}{75}\cdot\dfrac{26}{8}=\dfrac{13}{12}\)

b: Để A<x<B thì -11/12<x<13/12

mà x là số nguyên

nên \(x\in\left\{0;1\right\}\)

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