\(42-\left(2x+32\right)+12:2=6\)

\(\Rightarrow42-\left(2x+32\right)+12=12\)

\(\Rightarrow42-\left(2x+32\right)=0\)

\(\Rightarrow2x+32=42\)

\(\Rightarrow2x=10\)

\(\Rightarrow x=5\)

42-(2x+32)+12:2=6

(2x+32)+6=42-6

(2x-32)+6=36

(2x-32)=36-6

  2x-32=30

       2x=30+32

       2x=62

         x=62:2

         x=31

Tk mk nha

 xét 2A=2²+2³+2⁴+...+2¹¹

-A=2+2²+2³+......+2¹⁰

A=2¹¹-2

ta thấy 2/3 dư 2

          2²=4/3 dư 2

          2³=8/3 3 dư 2

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

2¹¹/3 dư 2

=>2¹¹-2laf 1 số chia hết cho 3

2A=2(2+2^2+2^3+2^4+...+2^8+2^9+2^10)

2A=2^2+2^3+2^4+2^5+...+2^9+2^10+2^11)

2A-A=(2^2+2^3+2^4+2^5+...+2^9+2^10+2^11)-(2+2^2+2^3+2^4+...+2^8+2^9+2^10)

A=2^11-2

A=2046

Mà 2046 chia hết cho 3

Vậy A chia hết cho 3 

Điều phải chứng minh

7 . ( x - 9 ) =35

x-9=35:7

x-9=5

x=5+9

x=14

7 . ( x - 9 ) = 35

x - 9 = 35 : 7

x - 9 = 5

x = 5 + 9

x = 14

(x-2)²=(x-2)⁴

2=4

\(\left(x-2\right)^4-\left(x-2\right)^2=0\)

\(\left(x-2\right)^2.\left[\left(x-2\right)^2-1\right]=0\)

\(=>\orbr{\begin{cases}\left(x-2\right)^2=0\\\left(x-2\right)^2-1=0\end{cases}}\)

\(=>\orbr{\begin{cases}x=2\\x=3\end{cases}}\)

\(\left(x-1\right)\left(x-7\right)=0\)

\(\Leftrightarrow\orbr{\begin{cases}x-7=0\\x-1=0\end{cases}}\)

\(\Leftrightarrow\orbr{\begin{cases}x=7\\x=1\end{cases}}\)

(x-1).(x-7)=0

\(\orbr{\begin{cases}x-1=0\\x-7=0\end{cases}}\)\(\Leftrightarrow\)\(\orbr{\begin{cases}x=1\\x=7\end{cases}}\)

Vậy x \(\in\){1;7}

HOK TỐT

a) 39 + ( 31 - x ) = 70

31 - x = 39 + 70

31 - x = 109

=> x = 31 - 109

=> x = -78

39 + ( 31 - x ) = 70

         31 - x = 70 - 39

         31 - x = 31

               x = 31 - 31

               x = 0

5^x-2-3²=2⁴-(2¹²-2¹²)

5^x-2-3²=2⁴-0

5^x-2=2⁴+3²

5^x-2=5⁶

x-2=6

x=8