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 !

ta có: 2005 + 53-54-55+56-57-58-59+60+61-62-63+64+65-66-67+68+69

      = 2005+(53-54-55+56)+(57-58-59+60)+(61-62-63+64)+(65-66-67+68)+69

      = 2005+0+0+0+0+69

      = 2005+69

      = 2074

h

\(\left(\dfrac{x+1}{55}+\dfrac{x+2}{56}+\dfrac{x+3}{57}+\dfrac{x+4}{58}\right)-4=0\)

<=>\(\dfrac{x+1}{55}+\dfrac{x+2}{56}+\dfrac{x+3}{57}+\dfrac{x+4}{58}=4\)

<=>\(\dfrac{x+1}{55}-1+\dfrac{x+2}{56}-1+\dfrac{x+3}{57}+\dfrac{x+4}{58}-1=4-4\)

<=>\(\dfrac{x+1}{55}-\dfrac{55}{55}+\dfrac{x+2}{56}-\dfrac{56}{56}+\dfrac{x+3}{57}-\dfrac{57}{57}+\dfrac{x+4}{58}-\dfrac{58}{58}=0\)

<=>\(\dfrac{x-54}{55}+\dfrac{x-54}{56}+\dfrac{x-54}{57}+\dfrac{x-54}{58}=0\)

<=>\(\left(x-54\right)\left(\dfrac{1}{55}+\dfrac{1}{56}+\dfrac{1}{57}+\dfrac{1}{58}\right)=0\)

<=>x-54=0

<=>x=54

vậy phương trình có tập nghiệm là S={54}

=>\(\left(\dfrac{x+1}{55}-1\right)+\left(\dfrac{x+2}{56}-1\right)+\left(\dfrac{x+3}{57}-1\right)+\left(\dfrac{x+4}{58}-1\right)=0\)

=>x-54=0

=>x=54

1711

cho xin vài tick nhé các bạn    Thanks

 Số các số hạng là:

(58 - 1) + 1 = 58 (số)

Tổng là:

(58 + 1) x 58 : 2 = 1711   

B=1-4+4²-4³+...+4⁵⁶-4⁵⁷

=> -4B=-4+4²-4³+4⁴+...+4⁵⁶-4⁵⁷+4⁵⁸

=> B-(-4B) = 1+(-4+4²-4³+...+4⁵⁶-4⁵⁷)-(-4+4²-4³+4⁴+...+4⁵⁶-4⁵⁷+4⁵⁸)

<=> 5B=1-4⁵⁸

=> \(B=\frac{1-4^{58}}{5}\)

Lời giải chi tiết:

\(57,65m.......576,5dm\)

a) (-30) + 17 + (-70) + 33

=[-30+(-70)]+(17+33)

=-100+50

=-50

b) 56 - 15 - 6 + (-75)

=(56-6)-[15+(-75)]

=50-(-60)

=110

c) (157 - 56) - (44 + 57)

=157-56-44+57

=(-157-57)-(56+44)

=-100-100

=-200

d) (-24 + 174) - (174 - 24)

=-24+174-174+24

=(-24+24)+(174-174)

=0+0

=0

e) (35 - 47) - (-123 + 35) + [20 - (123 - 57)]

=35-47+123-35+[20-123+57]

=35-47+123-35+20-123+57

=(35-35)+(123-123)+(-47+57)

=0+0+(-10)

=-10

Bn sai ở câu e dòng 4

Còn câu f thì sao?

=100+3.{144-4.[5-0]+1}

=100+3.{140.5+1}

=100+3.{700+1}

=100+3.701

=100+2103

=2203

\(10^2+3\cdot\left\{12^2-4\cdot\left[\left(5^7:5^6-5^0\right)\right]+1\right\}\)

\(=100+3\cdot\left\{144-4\cdot\left[\left(5-1\right)+1\right]\right\}\)

\(=100+3\cdot\left[144-4\cdot\left(4+1\right)\right]\)

\(=100+3\cdot\left(144-4\cdot5\right)\)

\(=100+3\cdot\left(144-20\right)\)

\(=100+3\cdot124\)

\(=100+372\)

\(=472\)

Cho t bt đáp án của ai đúng đi