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

\(\frac{\text{A=1,11+0,19-13.2}}{2,06+0,54}\) -(\(\frac{1}{2}\)

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

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

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

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

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: \(=2016+\dfrac{\dfrac{1}{5}+\dfrac{3}{8}+\dfrac{5}{11}}{-\dfrac{3}{10}+\dfrac{9}{10}-\dfrac{15}{22}}=2016+\dfrac{453}{440}:\dfrac{-9}{110}\)

\(=2016-\dfrac{151}{12}=\dfrac{24343}{12}\)

b: \(=\dfrac{1,3-13.2}{2.6}-\dfrac{5}{6}:2\)

\(=\dfrac{-119}{26}-\dfrac{5}{12}=\dfrac{-779}{156}\)

c: \(=15\left(-1-\dfrac{5}{7}-\dfrac{2}{7}\right)+\left(-105\right)\cdot\dfrac{1}{105}\)

\(=-30-1=-31\)