\(67-\left[8+7.3^2-24:6+\left(9-7\right)^3\right]:15\\ =67-\left[8+7.9-4+2^3\right]:15\\ =67-\left[8+63-4+8\right]:15\\ =67-75:15\\ =67-5=62\)

cảm ơn bạn Nguyễn Hòang tùng nhé

a) \(\left(-16\right).1253.\left(-8\right).\left(-4\right).\left(-3\right)\)

\(=16.1253.8.4.3>0\)

b) \(13.\left(-24\right).\left(-15\right).\left(-8\right).4< 0\) 

Phương pháp : ta nhóm 2 số âm với nhau rồi nhân vào thì làm mất dấu \("-"\)                       

1

(-57).(67-34)-67.(34-57)

=-57.67+57.34-67.34+67.57

=(-57.67+67.57)+57.34-67.34

=0+34(57-67)

=34.(-10)

=-340

2

a)(-16).(-8).(-4).(-3)>0 ; 1253>0 => (-16).1253.(-8).(-4).(-3)>0

b)13.4>0 ; (-24).(-15).(-8)<0=>13.(-24).(-15).(-8).4<0

Ta có:

67 - [ 8 + 7 . 3² - 24 : 6 + ( 9 - 7 )³ ] : 15 

= 67 - (8 + 7.9 - 4 + 8) : 15

= 67 - 75 : 15

= 67 - 5

= 62

đổi ra thì ta có:

67-[8+7.9-24:6+(9-7)^3]:15

=67-[8+63-4+2^3]:15

=67-[8+63- 4+8]:15

=67-[71-4+8]:15

=67-75:15

=67-5

=62

67 - [ 8 + 7 . 3² - 24 : 6 + ( 9 - 7 )³ ] : 15

= 67 - [ 8 + 7 . 9 - 4 + 2³ ] : 15

= 67 - [ 8 + 63 - 4 + 8 ] : 15

= 67-[71-4+8]:15

= 67 - [ 67 + 8 ] : 15

= 67 - 75 : 15

= 67 - 5

= 62

\(67-\left[8+7.3^2-24:6+\left(9-7\right)^3\right]:15\)

\(67-\left[8+7.9-4+2^3\right]:15\)

\(67-\left[8+63-4+8\right]:15\)

\(67-\left[71-11\right]:15\)

\(67-60:15\)

\(67-4\)

\(=63\)

67-\([8+7\times3^2-24:6+\left(9-7\right)^3]:15\)

\(\Leftrightarrow\)67-\([8+63-4+8]\) : 15 

\(\Leftrightarrow\)(67- 75):15

\(\Leftrightarrow\)-8 : 15

\(\Leftrightarrow\frac{-8}{15}\)

bạn chép lại đề đi, dấu [ ở chỗ nào đấy

67-[8+7.3²-24:6+(9-7)³]:15

=67-[8+7.9-4+8]:15

=67-[8+63-4+8]:15

=67-75:15

=67-5

=62

67-[8+7.9-24:6+2^3:15

=67-[8+63-4+8/15

=67-5

=62

67-[8+7.\(3^2\)-24:6+\(\left(9-7\right)^3\)]:15

=67-[8+63-4+8]:15

=67-75:15

=67-5

=62

67 -[ 8 + 7 x 3² - 24: 6 + (9 - 7 )³] :15

= 67 -[ 8 + 7 . 3² - 24 : 6 + 2³ ] : 15

= 67 - [ 8 + 7 . 9 - 24 : 6 + 8 ] : 15

= 67 - [ 71 - 12 ] : 15

= 67 - 59 : 15

= 8 : 15

=> ko thực hiện đc .

67-[8+7.3²-24:6+(9-7)³]:15

=67-[8+7.9-24:6+8]:15

=67-[8+63-4+8]:15

=67-75:15

=67-5

=62

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 !