P= 7 + \(7^2+7^3+7^4+...+7^{2016}\)

=\(\left(7+7^2+7^3+7^4\right)+\left(7^5+7^6+7^7+7^8\right)+...+\left(7^{2013}+7^{2014}+7^{2015}+7^{2016}\right)\)

=\(\left(7+7^2+7^3+7^4\right)+7^4\left(7+7^2+7^3+7^4\right)+...+7^{2012}\left(7+7^2+7^3+7^4\right)\)=2800+\(7^4\).2800+..+\(7^{2012}\).2800 \(⋮\) \(20^2\) ( Vì 2800 \(⋮\)\(20^2\))

=> P\(⋮\) \(20^2\)

Câu đúng là: b, c ,d ,e ,n , q

Ta có : \(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\)

=> \(\dfrac{x}{y}+\dfrac{y}{z}+\dfrac{z}{x}=\dfrac{y}{x}+\dfrac{z}{y}+\dfrac{x}{z}\)=\(\dfrac{x+y+z}{x+y+z}=1\)

=> x=y=z

TH1: x=y=z=1=>x+y+z=3( thỏa )

TH2: x,y,z >1=>x+y+z>3 ( vô lý)

Vậy x=y=z=1

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

=> \(\left\{{}\begin{matrix}\left|x-\dfrac{1}{2}\right|=0\\\left|y+\dfrac{2}{3}\right|=0\\\left|x^2+xz\right|=0\end{matrix}\right.\)

* \(\left|x-\dfrac{1}{2}\right|\)=0

=> x- \(\dfrac{1}{2}\)=0

=>x= \(\dfrac{1}{2}\)

* \(\left|y+\dfrac{2}{3}\right|\)=0

=> y+\(\dfrac{2}{3}\)=0

=>y = \(\dfrac{-2}{3}\)

* \(\left|x^2+xz\right|\)=0

mà x= \(\dfrac{1}{2}\) (cmt)

=> \(\left(\dfrac{1}{2}\right)^2+\dfrac{1}{2}z=0\)

=>\(\dfrac{1}{2}z=\dfrac{-1}{4}\)

=>z= \(\dfrac{-1}{2}\)

Vậy x,y,z= \(\dfrac{1}{2};\dfrac{-2}{3};\dfrac{-1}{2}\)

Tính  lim x → - 1   x 2 + 2 x + 1 2 x 3 + 2 .

A.  - ∞  

B. 0 

C.  1 2  

D.  + ∞  

a, \(\sqrt{x+1}=7\\ \Rightarrow x+1=49\\ \Rightarrow x=48\)

b,TH1:

\(\left\{{}\begin{matrix}x-2>0\\x +\dfrac{2}{3}>0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x>2\\x>\dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow x>2\)

TH2:

\(\left\{{}\begin{matrix}x-2< 0\\x+\dfrac{2}{3}< 0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x< 2\\x< \dfrac{-2}{3}\end{matrix}\right.\Leftrightarrow x< \dfrac{-2}{3}\)

=> Vậy 2<x< \(\dfrac{-2}{3}\)

c, TH1:

\(\dfrac{2}{3}x-1=0\\ \Rightarrow\dfrac{2}{3}x=1\\ \Rightarrow x=\dfrac{3}{2}\)

TH2:

\(\dfrac{3}{4}x+\dfrac{1}{2}=0\\ \Rightarrow\dfrac{3}{4}x=\dfrac{-1}{2}\\ \Rightarrow x=\dfrac{-2}{3}\)

Vậy x = \(\dfrac{3}{2};\dfrac{-2}{3}\)