=21 .[ 2232 : 8 - 180 ] + 21

=21.99 +21

=2100

\(21\cdot\left[\left(1245+987\right):2^3-15\cdot12\right]+21\)

\(=21\left(279-180+1\right)\)

\(=21\cdot100=2100\)

a) ( 12 + 21 - 23 ) - ( 23 - 21 + 10 )

  = 12 + 21 - 23 - 23 - 21 + 10

  = ( 12 + 10 ) - ( 21 - 21 ) - ( 23 - 23 )

  =        22      -        0        -        0

  =                          22

b) ( 55 + 45 + 15 ) - ( 15 - 55 + 45 )

  = 55 + 45 + 15 - 15 - 55 + 45

  = ( 55 + 45 ) + ( 55 + 45 ) + ( 15 - 15)

  =       100      +      100      +      0

  =                          200

\(a)\)\(-\left(12+21-23\right)-\left(23-21+10\right)\)

 \(=-12-21+23-23+21-10\)

 \(=\left(-21+21\right)+23-23-12-10\)

 \(=0+0-12-10\)

 \(=-22\)

\(b)\)\(\left(55+45+15\right)-\left(15-55+45\right)\)

  \(=55+45+15-15+55-45\)

  \(=\left(15-15\right)+\left(45-45\right)+55+55\)

  \(=0+0+55+55\)

  \(=110\)

a) = 22

b) =200

Bài 1: 

\(A=\dfrac{-1}{3}+1+\dfrac{1}{3}=1\)

\(B=\dfrac{2}{15}+\dfrac{5}{9}-\dfrac{6}{9}=\dfrac{2}{15}-\dfrac{1}{9}=\dfrac{18-15}{135}=\dfrac{3}{135}=\dfrac{1}{45}\)

\(C=\dfrac{-1}{5}+\dfrac{1}{4}-\dfrac{3}{4}=\dfrac{-1}{5}-\dfrac{1}{2}=\dfrac{-7}{10}\)

Bài 2: 

a: \(=\dfrac{1}{5}+\dfrac{1}{2}+\dfrac{2}{5}-\dfrac{3}{5}+\dfrac{2}{21}-\dfrac{10}{21}+\dfrac{3}{20}\)

\(=\left(\dfrac{1}{5}+\dfrac{2}{5}-\dfrac{3}{5}\right)+\left(\dfrac{2}{21}-\dfrac{10}{21}\right)+\left(\dfrac{1}{2}+\dfrac{3}{20}\right)\)

\(=\dfrac{-8}{21}+\dfrac{13}{20}=\dfrac{113}{420}\)

b: \(B=\dfrac{21}{23}-\dfrac{21}{23}+\dfrac{125}{93}-\dfrac{125}{143}=\dfrac{6250}{13299}\)

Bài 3:

\(\dfrac{7}{3}-\dfrac{1}{2}-\left(-\dfrac{3}{70}\right)=\dfrac{7}{3}-\dfrac{1}{2}+\dfrac{3}{70}=\dfrac{490}{210}-\dfrac{105}{210}+\dfrac{9}{210}=\dfrac{394}{210}=\dfrac{197}{105}\)

\(\dfrac{5}{12}-\dfrac{3}{-16}+\dfrac{3}{4}=\dfrac{5}{12}+\dfrac{3}{16}+\dfrac{3}{4}=\dfrac{20}{48}+\dfrac{9}{48}+\dfrac{36}{48}=\dfrac{65}{48}\)

Bài 4:

 \(\dfrac{3}{4}-x=1\)

\(\Rightarrow-x=1-\dfrac{3}{4}\)

\(\Rightarrow x=-\dfrac{1}{4}\)

Vậy: \(x=-\dfrac{1}{4}\)

\(x+4=\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{1}{5}-4\)

\(\Rightarrow x=-\dfrac{19}{5}\)

Vậy: \(x=-\dfrac{19}{5}\)

\(x-\dfrac{1}{5}=2\)

\(\Rightarrow x=2+\dfrac{1}{5}\)

\(\Rightarrow x=\dfrac{11}{5}\)

Vậy: \(x=\dfrac{11}{5}\)

\(x+\dfrac{5}{3}=\dfrac{1}{81}\)

\(\Rightarrow x=\dfrac{1}{81}-\dfrac{5}{3}\)

\(\Rightarrow x=-\dfrac{134}{81}\)

Vậy: \(x=-\dfrac{134}{81}\)

\(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}\) \(\left(a\ne0\right)\)

Tại a = 12 biểu thức \(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{12}\right):\dfrac{4}{5}=\dfrac{5}{4}-\dfrac{11}{24}:\dfrac{4}{5}=\dfrac{5}{4}-\dfrac{11}{24}.\dfrac{5}{4}=\dfrac{5}{4}-\dfrac{55}{96}=\dfrac{65}{96}\)

Để \(A=\dfrac{15}{23}< =>\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{15}{23}\)

\(\Leftrightarrow\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{55}{92}< =>\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{11}{23}< =>\dfrac{1}{a}=\dfrac{19}{184}< =>a=\dfrac{184}{19}\)

Thay \(a=12\) vào A ta có:

\(A=\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{12}\right):\dfrac{4}{5}=\dfrac{65}{96}\)

Vậy:

____________________

Ta có:

\(A=\dfrac{15}{23}\) khi \(\dfrac{5}{4}-\left(\dfrac{3}{8}+\dfrac{1}{a}\right):\dfrac{4}{5}=\dfrac{15}{23}\)

\(\Rightarrow\left(\dfrac{3}{8}+\dfrac{1}{a}\right)\cdot\dfrac{5}{4}=\dfrac{5}{4}-\dfrac{15}{23}\)

\(\Rightarrow\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{55}{92}:\dfrac{5}{4}\)

\(\Rightarrow\dfrac{3}{8}+\dfrac{1}{a}=\dfrac{11}{23}\)

\(\Rightarrow\dfrac{1}{a}=\dfrac{11}{184}\)

\(\Rightarrow a=\dfrac{1\cdot184}{11}=\dfrac{184}{11}\)

giúp tui đi

A = - \(\dfrac{1}{21}\) - \(\dfrac{1}{28}\)

A =- \(\dfrac{4}{84}\)- \(\dfrac{3}{84}\)

A = -\(\dfrac{7}{84}\)  

A = - \(\dfrac{1}{12}\)

B = - \(\dfrac{8}{18}\) - \(\dfrac{15}{27}\)

B = - \(\dfrac{4}{9}\) - \(\dfrac{5}{9}\)

B = -1

C = -  \(\dfrac{5}{12}\) + 0,75

C =  -\(\dfrac{5}{12}\) + \(\dfrac{3}{4}\)

C = - \(\dfrac{5}{12}\) + \(\dfrac{9}{12}\)

C = \(\dfrac{4}{12}\)

C= \(\dfrac{1}{3}\)

D = 3,5 - { - \(\dfrac{2}{7}\))

D = \(\dfrac{7}{2}\) + \(\dfrac{2}{7}\)

D = \(\dfrac{49}{14}\) + \(\dfrac{4}{14}\)

D = \(\dfrac{53}{14}\) 

=(4x0.25)x(2x0.5)

=1x1=1