B1: ĐXXĐ: \(x\ne\pm2;x\ne-1\)

\(=\left(\dfrac{x-2}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}+\dfrac{x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)

\(=\left(\dfrac{x-2-2x-2+x}{\left(x+2\right)\left(x-2\right)}\right):\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)

\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}:\dfrac{-6\left(x+2\right)}{\left(x-2\right)\left(x+1\right)}\)

\(=\dfrac{-4}{\left(x+2\right)\left(x-2\right)}.\dfrac{\left(x-2\right)\left(x+1\right)}{-6\left(x+2\right)}=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}\)

b, \(A=\dfrac{2\left(x+1\right)}{3\left(x+2\right)^2}>0\)

\(\Leftrightarrow2x+2>0\) (vì \(3\left(x+2\right)^2\ge0\forall x\))

\(\Leftrightarrow x>-1\).

-Vậy \(x\in\left\{x\in Rlx>-1;x\ne2\right\}\) thì \(A>0\).

a, \(\frac{1}{1\cdot2\cdot3}+\frac{1}{2\cdot3\cdot4}+...+\frac{1}{x\cdot\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)

\(=\frac{1}{1\cdot2}-\frac{1}{2\cdot3}+\frac{1}{2\cdot3}-\frac{1}{3\cdot3}+...+\frac{1}{x\cdot\left(x+1\right)}-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)

\(=1-\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2018}{2019}\)

\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=1-\frac{2018}{2019}\)

\(\Rightarrow\frac{1}{\left(x+1\right)\cdot\left(x+2\right)}=\frac{2019}{2019}-\frac{2018}{2019}=\frac{1}{2019}\)

Đến đây bn tự tính nhé !!

Nhưng \(\frac{1}{1.2.3}\)ko bằng \(\frac{1}{1.2}-\frac{1}{2.3}\)ạ

Bn có thể suy nghĩ lại giúp mik đc ko????

- Ấy chết mk quên, bn đưa \(\frac{1}{2}\cdot\left(...\right)\)

Vào nữa nhé. Mk thử lại nek:

\(\frac{1}{1\cdot2\cdot3}=\frac{1}{6}\)và \(\frac{1}{2}\cdot\left(\frac{1}{1\cdot2}-\frac{1}{2\cdot3}\right)=\frac{1}{2}\cdot\left(\frac{1}{2}-\frac{1}{6}\right)\)

\(=\frac{1}{2}\cdot\left(\frac{3}{6}-\frac{1}{6}\right)=\frac{1}{2}\cdot\frac{2}{6}=\frac{2}{12}=\frac{1}{6}\)nhé bn !!

a: \(A=\dfrac{x-2-2x-4+x}{\left(x-2\right)\left(x+2\right)}\cdot\dfrac{-\left(x-2\right)\left(x+1\right)}{6\left(x+2\right)}\)

\(=\dfrac{-6}{\left(x+2\right)}\cdot\dfrac{-\left(x+1\right)}{6\left(x+2\right)}=\dfrac{\left(x+1\right)}{\left(x+2\right)^2}\)

b: A>0

=>x+1>0

=>x>-1

c: x^2+3x+2=0

=>(x+1)(x+2)=0

=>x=-2(loại) hoặc x=-1(loại)

Do đó: Khi x^2+3x+2=0 thì A ko có giá trị

câu 1

A=-1

câu 2

\(\frac{x+1}{2}=\frac{8}{x+1}\)

\(\Rightarrow\left(x+1\right).\left(x+1\right)=8.2\)

\(\left(x+1\right).\left(x+1\right)=16\)

\(\left(x+1\right)^2=16\)

\(\Rightarrow x+1=4\)

vậy x=3

Câu 1:

Sai bét choét ...

Câu 2:

Đúng ròi

a,\(A=\left(\frac{2x-x^2}{2\left(x^2+4\right)}-\frac{2x^2}{\left(x^2+4\right)\left(x-2\right)}\right)\left(\frac{2x+x^2\left(1-x\right)}{x^3}\right)\left(ĐKXĐ:x\ne2;x\ne0\right)\)

\(A=\frac{\left(2x-x^2\right)\left(x-2\right)-4x^2}{2\left(x^2+4\right)\left(x-2\right)}.\frac{-x^3+x^2+2x}{x^3}\)

\(=\frac{-x^3-4x}{2\left(x^2+4\right)\left(x-2\right)}.\frac{x^2-x-2}{-x^2}\)

\(=\frac{-x\left(x^2+4\right)}{2\left(x^2+4\right)\left(x-2\right)}.\frac{\left(x-2\right)\left(x+1\right)}{-x^2}=\frac{x+1}{2x}\)

b, \(A=x\Leftrightarrow\frac{x+1}{2x}=x\Rightarrow2x^2=x+1\Leftrightarrow2x^2-x-1=0\)

\(\Leftrightarrow\left(2x+1\right)\left(x-1\right)=0\Leftrightarrow\orbr{\begin{cases}x=-\frac{1}{2}\\x=1\end{cases}}\)(thỏa mãn điều kiện)

c, \(A\in Z\Leftrightarrow\frac{x+1}{2x}\in Z\Leftrightarrow x+1⋮\left(2x\right)\)

\(\Leftrightarrow2x+2⋮2x\Leftrightarrow2⋮2x\Leftrightarrow1⋮x\Leftrightarrow x=\pm1\) (thỏa mãn ĐKXĐ)

x+(x+1)+(x+2)+(x+3)+....+(x+2017)=20172018

\(\left(x+x+x+...+x\right)+\left(1+2+3+....+2017\right)=20172018\)

\(x.2018+\left(1+2017\right).2017:2=20172018\)

\(x.2018+2018.2017:2=20172018\)

\(x.2018+4070306:2=20172018\)

\(x.2018+2035153=20172018\)

\(x.2018=18136865\)

\(x=\frac{18136865}{2018}\)

\(x+\left(x+1\right)+\left(x+2\right)+\left(x+3\right)+...+\left(x+2017\right)=20172018\)

\(\Rightarrow x+\left(x+x+x+...+x\right)+\left(1+2+3+...+2017\right)=20172018\)

\(\text{Đặt}\) \(A=x+\left(x+x+x+...+x\right)\)

             \(B=1+2+3+...+2017\)

Ta có : 

Số số hạng của B là : \(\left(2017-1\right)\div1+1=2017\left(\text{số hạng }\right)\)

Tổng của B là           : \(\frac{\left(1+2017\right).2017}{2}=2035153\)

Vì B có 2017 số hạng nên A có :  \(x+2017x\)\(=2018x\)

\(\Rightarrow2018x+2035153=20172018\)

\(\Rightarrow2018x=20172018-2035153\)

\(\Rightarrow2018x=18136865\)

\(\Rightarrow x=\frac{18136865}{2018}=8987.544599\)

Vậy x = 8987.544599