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51)

52)a) 14 . 15 = (14 : 2) . ( 50 . 2)                         16 . 25 = (16: 4 ) . ( 25 . 4)

                      = 7 . 100                                                     = 4 . 100
                      = 700                                                          = 400

( x + 15 ) - 97 = 215

  x + 15          = 215 + 97

  x + 15          = 312

   x                 = 312 - 15

   x   = 297

A

<=>x+15=215+97=312

<=>x      =312-15

<=>x      =297

B

<=>2x=238-84=154

<=>x  =154:2

<=>x  =77

C

<=>x-93=7

<=>x     =7+93

<=>x     =100

P=2.4.6.8.16.32.64.128.256.512.1024.2^52

=2^1.2^2.2^3. .... .2^10.2^52

=2^(1+2+3+...+10+52)=2^107

11: \(27+12+\left(-17\right)+\left(-12\right)\)

\(=\left(27-17\right)+\left(12-12\right)\)

=10+0

=10

12: \(-564-\left(324-564-224\right)\)

\(=-564-\left(-564+100\right)\)

\(=-564+564-100=-100\)

13: \(2^3\cdot3^2-7^2-2022^0\)

\(=8\cdot9-49-1\)

=72-50

=22

14: \(236\cdot145+236\cdot856-236\)

\(=236\cdot145+236\cdot856-236\cdot1\)

\(=236\left(145+856-1\right)\)
\(=236\cdot1000=236000\)

15: \(-\left(-357\right)+27+\left(-357\right)+\left(-32\right)\)

\(=357+27-357-32\)

=27-32

=-5

16: \(275\cdot48+22\cdot275+175\left(-70\right)\)

\(=275\left(48+22\right)-175\cdot70\)

\(=275\cdot70-175\cdot70\)

\(=70\left(275-175\right)=70\cdot100=7000\)

17: \(\dfrac{\left(23\cdot36-17\cdot36\right)}{36}\)

\(=\dfrac{36\left(23-17\right)}{36}\)

=23-17

=6

18: \(3\cdot5^3-27:3^2+5^3\cdot4-18:3^2\)

\(=3\cdot5^3+4\cdot5^3-27:9-18:9\)

\(=5^3\left(3+4\right)-3-2\)

\(=125\cdot7-5=875-5=870\)

19: \(-87+\left(-12\right)-\left(-487\right)+512\)

\(=-87-12+487+512\)

\(=\left(487-87\right)+\left(512-12\right)\)

=400+500

=900

20: \(25\cdot57+75+33\cdot25+75\cdot89\)

\(=25\cdot57+25\cdot33+75+75\cdot89\)

\(=25\left(57+33\right)+75\left(1+89\right)\)

\(=25\cdot100+100\cdot75\)

\(=100\left(25+75\right)=100\cdot100=10000\)

Lời giải:
Ta có:

$10\equiv -1\pmod {11}$

$\Rightarrow 10^{2022}\equiv (-1)^{2022}\equiv 1\pmod {11}$

$\Rightarrow A=10^{2022}-1\equiv 1-1\equiv 0\pmod {11}$

Vậy $A\vdots 11$

ok

A= 10^2022-1

Ta có thể thấy 10^2022=100000000...........0000000000 

 10000000.......0000000000-1 thì lúc nnày tổng bằng

9999999999999999........................999999999999999999999

mà 99999999999999999999999....................9999999999999999999chia hết cho 11 nên tổng này chia hết cho 11

x-1=3 hoặc x-1=-3

=>x=4         => x=-2

............................

học tốt!!!!!!!!!!!!!!

\(\left|x-1\right|=3\)

\(\Leftrightarrow\orbr{\begin{cases}x-1=3\\x-1=-3\end{cases}\Leftrightarrow\orbr{\begin{cases}x=4\\x=-2\end{cases}}}\)

Vậy x=4; x=-2

Giải:

a)\(\dfrac{3}{4}x-\dfrac{1}{3}=\dfrac{-5}{6}\) 

           \(\dfrac{3}{4}x=\dfrac{-5}{6}+\dfrac{1}{3}\) 

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

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

             \(x=\dfrac{-2}{3}\) 

b)\(\left(2\dfrac{4}{5}x-0,2\right):\dfrac{4}{5}=\dfrac{7}{8}\) 

       \(\dfrac{14}{5}x-\dfrac{1}{5}=\dfrac{7}{8}.\dfrac{4}{5}\) 

        \(\dfrac{14}{5}x-\dfrac{1}{5}=\dfrac{7}{10}\) 

         \(\dfrac{14}{5}x=\dfrac{7}{10}+\dfrac{1}{5}\) 

          \(\dfrac{14}{5}x=\dfrac{9}{10}\) 

              \(x=\dfrac{9}{10}:\dfrac{14}{5}\) 

             \(x=\dfrac{9}{28}\) 

c) \(\dfrac{1}{4}+\dfrac{1}{3}:\left|2x-1\right|=\dfrac{11}{12}\) 

            \(\dfrac{1}{3}:\left|2x-1\right|=\dfrac{11}{12}-\dfrac{1}{4}\) 

            \(\dfrac{1}{3}:\left|2x-1\right|=\dfrac{2}{3}\) 

                  \(\left|2x-1\right|=\dfrac{1}{3}:\dfrac{2}{3}\) 

                  \(\left|2x-1\right|=\dfrac{1}{2}\) 

⇒2x-1=\(\dfrac{1}{2}\) hoặc 2x-1=\(\dfrac{-1}{2}\) 

        x=\(\dfrac{3}{4}\) hoặc x=\(\dfrac{1}{4}\)

câu a nha \(\dfrac{3}{4}x-\dfrac{1}{3}=-\dfrac{5}{6}\)

\(\dfrac{3}{4}x=-\dfrac{5}{6}+\dfrac{1}{3}\)

\(\dfrac{3}{4}x=-\dfrac{3}{6}\)

\(x=-\dfrac{3}{6}:\dfrac{3}{4}\)

\(x=-\dfrac{6}{4}.\dfrac{4}{3}\)

\(x=-\dfrac{24}{12}=-2\)