300

300

= 300 

Bài 1: 

1) Ta có: \(\left(-12\right)+6\cdot\left(-3\right)\)

\(=-12-18\)

=-30

2) Ta có: \(\left(36-2020\right)+\left(2019-136\right)-27\)

\(=36-2020+2019-136-27\)

\(=1-100-27\)

\(=-126\)

3) Ta có: \(\left(144-97\right)-\left(244-197\right)\)

\(=144-97-244+197\)

\(=-100+100=0\)

4) Ta có: \(\left(-24\right)\cdot13-24\cdot\left(-3\right)\)

\(=-24\cdot13+24\cdot3\)

\(=24\cdot\left(-13+3\right)\)

\(=24\cdot\left(-10\right)=-240\)

5) Ta có: \(54+55+56+57+58-\left(64+65+66+67+68\right)\)

\(=54+55+56+57+58-64-65-66-67-68\)

\(=\left(54-64\right)+\left(55-65\right)+\left(56-66\right)+\left(57-67\right)+\left(58-68\right)\)

\(=\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)+\left(-10\right)\)

=-50

6) Ta có: \(24\cdot\left(16-5\right)-16\cdot\left(24-5\right)\)

\(=24\cdot16-24\cdot5-16\cdot24+16\cdot5\)

\(=-24\cdot5+16\cdot5\)

\(=5\cdot\left(-24+16\right)\)

\(=-5\cdot8=-40\)

7) Ta có: \(47\cdot\left(23+50\right)-23\cdot\left(47+50\right)\)

\(=47\cdot23+47\cdot50-23\cdot47-23\cdot50\)

\(=47\cdot50-23\cdot50\)

\(=50\cdot\left(47-23\right)\)

\(=50\cdot24=1200\)

8) Ta có: \(\left(-31\right)\cdot47+\left(-31\right)\cdot52+\left(-31\right)\)

\(=-31\cdot\left(47+52+1\right)\)

\(=-31\cdot100=-3100\)

Bài 2: 

1) Ta có: \(-17-\left(2x-5\right)=-6\)

\(\Leftrightarrow-17-2x+5+6=0\)

\(\Leftrightarrow-2x-6=0\)

\(\Leftrightarrow-2x=6\)

hay x=-3

Vậy: x=-3

2) Ta có: \(10-2\left(4-3x\right)=-4\)

\(\Leftrightarrow10-8+6x+4=0\)

\(\Leftrightarrow6x+6=0\)

\(\Leftrightarrow6x=-6\)

hay x=-1

Vậy: x=-1

3) Ta có: \(-12+3\left(-x+7\right)=-18\)

\(\Leftrightarrow-12-3x+21+18=0\)

\(\Leftrightarrow-3x+27=0\)

\(\Leftrightarrow-3x=-27\)

hay x=9

Vậy: x=9

4) Ta có: \(-45:\left[5\cdot\left(-3-2x\right)\right]=3\)

\(\Leftrightarrow5\cdot\left(-3-2x\right)=-15\)

\(\Leftrightarrow-2x-3=-3\)

\(\Leftrightarrow-2x=0\)

hay x=0

Vậy: x=0

5) Ta có: x(x+3)=0

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-3\right\}\)

6) Ta có: (x-2)(x+4)=0

\(\Leftrightarrow\left[{}\begin{matrix}x-2=0\\x+4=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=2\\x=-4\end{matrix}\right.\)

Vậy: \(x\in\left\{2;-4\right\}\)

7) Ta có: \(x\left(x+1\right)\left(x-3\right)=0\)

\(\Leftrightarrow\left[{}\begin{matrix}x=0\\x+1=0\\x-3=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=0\\x=-1\\x=3\end{matrix}\right.\)

Vậy: \(x\in\left\{0;-1;3\right\}\)

Bài 1: 

1) Ta có: (−12)+6⋅(−3)(−12)+6⋅(−3)

=−12−18=−12−18

=-30

2) Ta có: (36−2020)+(2019−136)−27(36−2020)+(2019−136)−27

=36−2020+2019−136−27=36−2020+2019−136−27

=1−100−27=1−100−27

=−126

Tớ chcs cậu học thật giỏi nha !

=5743 minh tin tuong minh 10000%

= 5743

Nhớ k cho mình nhé !  Thank you very much !

36 x 23 + 62 x 23 + 46 = 36 x 23 + 62 x 23 + 2 x 23 

                       = 23 x (36 + 62 + 2)

                   = 23 x 100 = 2300

45-58+(-45)+(+58)-3=45-58-45+58-3=(45-45)+(58-58)-3=-3

(-57)-(-12)+3-(-57)=-57+12+3+57=(-57+57)+12+3=15

-(-24)-(-30)+(-24)-30+8=24+30-24-30+8=(24-24)+(30-30)+8=8

-(-12)+(+19)-(+12)+8-19=12+19-12+8-19=(12-12)+(19-19)+8=8

47-(-53)+(-47)-(+53)=47+53-47-53=(47-47)+(53-53)=0

-(+28)+32-(-28)+(-34)=-28+32+28-34=(-28+28)-(34-32)=-2

15-(+72)+(-15)+(+72)=15-72-15+72=(15-15)+(72-72)=0

20-(+46)+(-25)-(-46)=20-46-25+46=(20-25)+(46-46)=-5

28-(-32)+(-28)-32+5=28+32-28-32+5=(28-28)+(32-32)+5=5

-75+(-49)+(+75)+49-12=-75-49+75+49-12=(-75+75)+(49-49)-12=-12

\(...=45-58-45+58-3=-3\)

\(...=-57+12+3+57=15\)

\(...=24+30-24-30+8=8\)

\(...=12+19-12+8-19=8\)

\(...=47+53-47-53=0\)

\(...=-28+32+28-34=-2\)

\(...=15-72-15+72=0\)

\(...=20-46-25+46=-5\)

\(=...28+32-28-32+5=5\)

\(...=-75-49+75+49-12=-12\)

`#3107.101107`

\(\left[2^3\cdot97+8\cdot75-\left(-47\right)\cdot2^3\right]\div10^2\)

\(=\left(2^3\cdot97+2^3\cdot75+47\cdot2^3\right)\div10^2\)

\(=\left[2^3\cdot\left(97+75-47\right)\right]\div10^2\)

\(=\left(2^3\cdot125\right)\div10^2\)

\(=\left(2^3\cdot5^3\right)\div10^2\)

\(=\left(5\cdot2\right)^3\div10^2\)

\(=10^3\div10^2\)

\(=10\)

\(\left(2^3.97+8.75-47.2^3\right):10^2\\ =\left[\left(97-47+75\right).2^3\right]:100\\ =\left[125.8\right]:100\\ =1000:100=10\)