giup mk

Hình như sai đề bài đó

P =2+2^2+2^3+2^4+2^5+2^6+2^7+2^8+2^9+2^10  

    = ( 2+2^2 ) + ( 2^3 + 2^4 ) + ( 2^5 + 2^6 ) + ( 2^7 + 2^8 ) + ( 2^9 + 2^10 )

   = 2.( 1+2 ) + 2^3.( 1+2 ) + 2^5.( 1+2 ) + 2^7.( 1+2 ) + 2^9.( 1 + 2 )

   = 2.3+2^3.3+2^5.3+2^7.3+2^9.3

   = ( 2+2^3+2^5+2^7+2^9).3 chia hết cho 3

=> P chia hết cho 3

= mấy hả bạn

Để căn thức có nghĩa\(\Leftrightarrow\left\{{}\begin{matrix}-\dfrac{2}{x+1}\ge0\\x+1\ne0\end{matrix}\right.\)\(\Leftrightarrow\left\{{}\begin{matrix}x+1\le0\\x+1\ne0\end{matrix}\right.\)\(\Leftrightarrow x+1< 0\Leftrightarrow x< -1\)

Vậy...

ĐKXĐ: x<-1

2x ( x + 3 ) + 3 ( 2 + x ) = 62

3x + 6x + 6 + 3x = 62

3x + 6x + 3x = 62 - 6

12x = 56

x = 56 : 12

x = 14/3

Vậy x = 14/3

Chắc sai =))

\(\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}=\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}\) \(\left(dkxd:x\ne-2\right)\)

\(\Leftrightarrow\dfrac{\left(x^2+x+1\right)\left(3x+1\right)}{x+2}-\dfrac{x\left(x^2+x+1\right)}{2\left(x+2\right)}=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left[2\left(3x+1\right)-x\right]=0\)

\(\Leftrightarrow\left(x^2+x+1\right)\left(6x+2-x\right)=0\)

Bỏ vế đằng trước \(x^2+x+1=0\) do vô nghiệm

\(\Leftrightarrow6x+2-x=0\)

\(\Leftrightarrow5x=-2\)

\(\Leftrightarrow x=-\dfrac{2}{5}\left(tmdk\right)\)

Vậy \(S=\left\{-\dfrac{2}{5}\right\}\)

\(\dfrac{\left(x^2+x+1\right).\left(3x+1\right)}{x+2}=\dfrac{\left(x^2+x+1\right).x}{2\left(x+2\right)}\)

hay \(\left(x^2+x+1\right).\dfrac{3x+1}{x+2}=\left(x^2+x+1\right).\dfrac{x}{2\left(x+2\right)}\)

3/5 x = 5/6

x = 5/6 : 3/5

x = 25/18

\(\dfrac{3}{5}:x=\dfrac{1}{3}+\dfrac{1}{2}\)

\(\dfrac{3}{5}:x=\dfrac{5}{6}\)

\(x=\dfrac{3}{5}:\dfrac{5}{6}\)

\(x=\dfrac{18}{25}\)

3/5:*=1/3+1/2

3/5:*=5/6

*=5/6:3/5

*=25/18