a) Đặt: \(A=1+2^2+2^3+...+2^{10}\)

\(\Rightarrow2A=2\left(1+2^2+2^3+...+2^9+2^{10}\right)\)

\(\Rightarrow2A=2+2^3+2^4+...+2^{10}+2^{11}\)

\(\Rightarrow2A-A=\left(2+2^3+2^4+...+2^{10}+2^{11}\right)-\left(1+2^2+2^3+...+2^{10}\right)\)

\(\Rightarrow A=\left(2^3-2^3\right)+\left(2^4-2^4\right)+...+\left(2-1\right)+\left(2^{11}-2^2\right)\)

\(\Rightarrow A=0+0+...+1+\left(2^{11}-2^2\right)\)

\(\Rightarrow A=1+2^{11}-2^2=1+2048-4=2045\)

Vậy: \(1+2^2+2^3+...+2^{10}=2045\)

b) 

a] \(60-3\left(x-1\right)=2^3\cdot3\)

\(\Rightarrow60-3\left(x-1\right)=24\)

\(\Rightarrow3\left(x-1\right)=36\)

\(\Rightarrow x-1=12\)

\(\Rightarrow x=13\)

b] \(\left(3x-2\right)^3=2\cdot2^5\)

\(\Rightarrow\left(3x-2\right)^3=2^6\)

\(\Rightarrow\left(3x-2\right)^3=\left(2^2\right)^3\)

\(\Rightarrow3x-2=2^2\)

\(\Rightarrow3x=6\)

\(x=2\)

c] \(5^{x+1}-5^x=500\)

\(\Rightarrow5^x\left(5-1\right)=500\)

\(\Rightarrow5^x\cdot4=500\)

\(\Rightarrow5^x=125\)

\(\Rightarrow5^x=5^3\)

\(\Rightarrow x=3\)

d] \(x^2=x^4\)

\(\Rightarrow x=x^2\)

\(\Rightarrow x-x^2=0\)

\(\Rightarrow x\left(1-x\right)=0\)

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

\(\Rightarrow\left[{}\begin{matrix}x=0\\x=1\end{matrix}\right.\)

giúp mình đi các bạn

Bài 2: 

a: \(\Leftrightarrow x=\dfrac{29}{60}\cdot\dfrac{-7}{5}=\dfrac{-203}{300}\)

b: \(\Leftrightarrow x\cdot\dfrac{2}{5}=\dfrac{29}{60}-\dfrac{3}{4}=\dfrac{29-45}{60}=\dfrac{-16}{60}=\dfrac{-8}{30}\)

\(\Leftrightarrow x=\dfrac{-8}{30}:\dfrac{2}{5}=\dfrac{-8\cdot5}{30\cdot2}=\dfrac{-40}{60}=-\dfrac{2}{3}\)

`3/4+2/5x=29/60`

`2/5x=29/60-3/4`

`2/5x=-4/15`

`x=-4/15:2/5`

`x=-2/3`

1. a, 15-(3+x)+2x=19
=> 15- 3 - x + 2x = 19
=> -x + 2x = 19 - 15 +3
=> ( -1 + 2 )x = 7
=> x = 7
b, (-58)-(19-58-x)=60
=> (-58)-(-39-x)=60
=> -58 +39 + x = 60
=> -19 + x = 60
=> x = 60 - (-19) = 79
c, 2(x-4)-3(x+5)=4
=> (2x - 8 )-(3x+15)= 4
=> 2x - 8 - 3x - 15 = 4
=> 2x - 3x = 4 + 15 + 8
=> - x = 27
=> x = -27
d, (2-3x)+(x-3)=(5x-1)-(5x-2)
=> 2 - 3x + x - 3 = 5x - 1 - 5x + 2
=> 2 - 3 - 3x + x = 5x - 5x - 1 +2
=> 2 - 3 - 3x + x = 1
=> - 3x + x = 1 - 2 + 3
=> -2x = 2
=> x= -1
=))))

2.

a,

A= -3752-(29-3632)-51

A= -3572-29+3632-51

A=(-3572+3632)-29-51

A=

1> a, 15 - ( 3 + x ) + 2 x = 19
=> 15 - 3 - x + 2x = 19
=> -x + 2x = 19 - 15 +3
=> ( -1 + 2 ) x = 7
=> x = 7
b, ( -58 ) - ( 19-58-x ) = 60
=> ( -58 ) - ( -39 – x ) = 60
=> -58 + 39 + x = 60
=> -19 + x = 60
=> x = 60 - ( -19 ) = 79
c, 2 ( x - 4 ) - 3 ( x+5 ) = 4
=> ( 2x - 8 ) - ( 3x + 15 ) = 4
=> 2x - 8 - 3x - 15 = 4
=> 2x - 3x = 4 + 15 + 8
=> - x = 27
=> x = -27
d, ( 2 - 3x ) + ( x - 3 ) = ( 5x - 1 ) - ( 5x – 2 )
=> 2 - 3x + x - 3 = 5x - 1 - 5x + 2
=> 2 - 3 - 3x + x = 5x - 5x - 1 +2
=> 2 - 3 - 3x + x = 1
=> - 3x + x = 1 - 2 + 3
=> -2x = 2
=> x= -1

a/|x|-2,5=27,5

=>|x|=27,5+2,5=30

=>x=30 hoặc x=-30

b/\(\dfrac{3}{4}+\dfrac{2}{5}.x=\dfrac{29}{60}\)

=>\(\dfrac{2}{5}.x\)=\(\dfrac{29}{60}-\dfrac{3}{4}\)=\(\dfrac{-4}{15}\)

=>x=\(\dfrac{-4}{15}:\dfrac{2}{5}\)=\(\dfrac{-2}{3}\)

c/(x-1)\(^5\)=-32

=>x-1=-2 vì (-2)\(^5\)=-32

=>x=-2+1=-1

d/\(\dfrac{4}{5}.x+0,5=4.5\)

=>\(\dfrac{4}{5}.x+0,5=20\)

=>\(\dfrac{4}{5}.x=20-0,5=19,5\)

=>\(x=19,5:\dfrac{4}{5}\)=\(\dfrac{195}{8}\)

5x-x=29-36:4

4x   =20

x     =20:4

x     =5

Vậy x=5

      5x-x=29-36:4
<=>4x=29-36:4

<=>4x=29-9

<=>4x=20

<=>x=5

Vậy...bla bla

\(5.x-x=29-36:4\)

\(5.x-x=29-9\)

\(5.x-x=20\)

\(5.x-x.1=20\)

\(\Rightarrow x.\left(5-1\right)=20\)

\(x.4=20\)

\(\Rightarrow x=5\)

5x-x - 29+ 9 bằng 0

4x-18 bằng 0

4x bằng 18 

x bằng \(\frac{9}{2}\)

5x - x = 29 - 36 : 4

5x - x = 29 - 9 = 20

4x      = 20

x        = 20 : 4

x        = 5

đùa thôi 4x - 20 bằng 0

x bằng 20 x bằng 5

5x - x = 29 - 36 : 4

4x = 29 - 9

4x = 20

x = 20 : 4

x = 5

5x - x = 29 - 36 : 4

4x      = 29 - 9

4x      = 20

 x       = 20 : 4

 x       = 5

chúc bạn học tốt ^ _ ^ 

5 x - x = 29 - 36 : 4

    4x   = 29 - 9

    4x   = 20

      x   = 20:4

      x   =   5