\(A=\left(\frac{1}{x+2}+\frac{1}{x-2}\right)\times\frac{x-2}{x}\)

a) ĐKXĐ : \(x\ne0;x\ne-2;x\ne2\)

b) Rút gọn:

\(A=\frac{x-2+x+2}{\left(x+2\right)\left(x-2\right)}\times\frac{x-2}{x}=\frac{2}{x+2}\)

c) Tìm x>0 để: \(A>\frac{1}{2}\Rightarrow\frac{2}{x+2}>\frac{1}{2}\)Do x>0 nên x + 2 >0 => x + 2 <4 =>x < 2.

Vậy, với x dương để A > 1/2 thì x<2 

d) A nguyên thì 2 chia hết cho x+2 hay x+2 thuộc U(2) = {-2;-1;1;2)

Có 4 giá trị nguyên của x là: -4; -3; -1; 0 để A có giá trị nguyên.

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)

\(1,\\ a,ĐK:\left\{{}\begin{matrix}x\ge0\\x+5\ge0\end{matrix}\right.\Leftrightarrow x\ge0\\ b,Sửa:B=\left(\sqrt{3}-1\right)^2+\dfrac{24-2\sqrt{3}}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+\dfrac{2\sqrt{3}\left(\sqrt{2}-1\right)}{\sqrt{2}-1}\\ B=4-2\sqrt{3}+2\sqrt{3}=4\\ 3,\\ =\left[1-\dfrac{\sqrt{x}\left(\sqrt{x}+1\right)}{1+\sqrt{x}}\right]\cdot\dfrac{\sqrt{x}-3+2-2\sqrt{x}}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\left(1-\sqrt{x}\right)\cdot\dfrac{-\sqrt{x}-1}{\left(1-\sqrt{x}\right)\left(\sqrt{x}-3\right)}-2\\ =\dfrac{-\sqrt{x}-1}{\sqrt{x}-3}-2=\dfrac{-\sqrt{x}-1-2\sqrt{x}+6}{\sqrt{x}-3}=\dfrac{-3\sqrt{x}+5}{\sqrt{x}-3}\)

\(\left(x+4\right)^2-81=0\Leftrightarrow\left(x+4\right)^2-9^2=0\)

\(\Leftrightarrow\left(x+4+9\right)\times\left(x+4-9\right)=0\)

\(\Leftrightarrow\left(x+13\right)\times\left(x-5\right)=0\)

\(\left[{}\begin{matrix}x+13=0\\x-5=0\end{matrix}\right.\Leftrightarrow\left[{}\begin{matrix}x=-13\\x=5\end{matrix}\right.\)

Bài 1 :

a) x={2,4}

b) x-1={-3,-2,-1,0,1,2,3,4}

=> x={-2,-1,0,1,2,3,4,5}

c) x+2={-7,-6,-5,-4}

=> x={-9,-8,-7,-6}

Bài 2 :

(x-3)(x+2)=0

=> x-3=0 => x=3

=> x+2=0 => x=-2

Vậy x=-2 hoặc x=3

BÀI 1

A) 3<X<5

=>X=4

B) -4<X+2<5

=>X-1\(\in\left(-3;-2;-1;0;1;2;3;4\right)\)

=> X-1=-3             => X-1=-2                  =>X-1=-1             =>X-1=0               => X-1=1

X=-2                              X=-1                        X=    0                 X=1                       X=2

=>X-1=2             => X-1=3             =>X-1=4

X=3                              X=4              X=5

C) -8<X+2<-3

=> X+2\(\in\left(-7;-6;-5;-4\right)\)

=> X+2=-7            =>X+2=-6          =>X+2=-5                =>X+2=-4

  X=-9                      X=-8                   X=-7                           X=-6

BÀI 2

\(\left(X-3\right).\left(X+2\right)=0\)

\(\Rightarrow X-3=X+2=O\)

\(TH1:X-3=0\)

              X=3

TH2: X+2=0

      X=-2

VẬY X=3 HOẶC X=-2

Cảm ơn câu trả lời của 2 bạn nhé ! :)