a ) \(A=\frac{1}{x-2}-\frac{1}{x+2}+\frac{x^2+4}{x^2-4}\)

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

\(=\frac{x^2+8}{x^2-4}\)

b ) \(A=\frac{x^2+8}{x^2-4}=\frac{\left(x^2-4\right)+12}{x^2-4}=1+\frac{12}{x^2-4}\)

Để \(A\in Z\Leftrightarrow12⋮x^2-4\)

\(x^2-4\inƯ\left(12\right)=\left\{-12;-6;-4;-2;-1;1;2;4;6;12\right\}\)

Xét từng thường hợp của x ta tìm đc : \(x=\left\{-4;0;4\right\}\)

\(\frac{1}{x-2}-\frac{1}{x+2}+\frac{x^2+4}{x^2-4}\)

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

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

=\(\frac{4}{x^2-2^2}+\frac{x^2+2^2}{x^2-2^2}\)

= \(\frac{4+x^2+2^2}{x^2-2^2}\)

\(A=\left(\frac{x+1}{x^3+1}-\frac{1}{x-x^2-1}-\frac{2}{x+1}\right)\div\left(\frac{x^2-2x}{x^3-x^2+x}\right)\)

a) ĐKXĐ : \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

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

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

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

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

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

b) \(\left|x-\frac{3}{4}\right|=\frac{5}{4}\)

<=> \(\orbr{\begin{cases}x-\frac{3}{4}=\frac{5}{4}\\x-\frac{3}{4}=-\frac{5}{4}\end{cases}}\Leftrightarrow\orbr{\begin{cases}x=2\left(loai\right)\\x=-\frac{1}{2}\left(nhan\right)\end{cases}}\)

Với x = -1/2 => \(A=\frac{-2\cdot\left(-\frac{1}{2}\right)}{-\frac{1}{2}+1}=2\)

c) Để A ∈ Z thì \(\frac{-2x}{x+1}\)∈ Z

=> -2x ⋮ x + 1

=> -2x - 2 + 2 ⋮ x + 1

=> -2( x + 1 ) + 2 ⋮ x + 1

Vì -2( x + 1 ) ⋮ ( x + 1 )

=> 2 ⋮ x + 1

=> x + 1 ∈ Ư(2) = { ±1 ; ±2 }

Các giá trị trên đều tm \(\hept{\begin{cases}x\ne-1\\x\ne2\end{cases}}\)

Vậy x ∈ { -3 ; -2 ; 0 ; 1 }

a) Điều kiện \(\begin{cases}x\ge0\\x-1\ne0\end{cases}\Leftrightarrow\begin{cases}x\ge0\\x\ne1\end{cases}}\)
Chú ý: x\(\ge0\) nên \(\sqrt{x}+1;4\sqrt{x}+4\) luôn khác 0

a) A xác định \(\Leftrightarrow\hept{\begin{cases}3x\ne0\\x+1\ne0\\2-4x\ne0\end{cases}\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne-1\\x\ne\frac{1}{2}\end{cases}}}\)

\(A=\left(\frac{x+2}{3x}+\frac{2}{x+1}-3\right):\frac{2-4x}{x+1}-\frac{3x+1-x^2}{3x}\)

\(A=\left[\frac{\left(x+2\right)\left(x+1\right)}{3x\left(x+1\right)}+\frac{2\cdot3x}{3x\left(x+1\right)}-\frac{3\cdot3x\left(x+1\right)}{3x\left(x+1\right)}\right]\cdot\frac{x+1}{2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{x^2+3x+2+6x-9x^2-9x}{3x\left(x+1\right)}\cdot\frac{x+1}{2\cdot\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{\left(-8x^2+2\right)\left(x+1\right)}{3x\left(x+1\right)2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2\left(1-4x^2\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2\left(1-2x\right)\left(1-2x\right)}{3x\cdot2\left(1-2x\right)}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{1+2x}{3x}-\frac{3x+1-x^2}{3x}\)

\(A=\frac{2x+1-3x-1+x^2}{3x}\)

\(A=\frac{x^2-x}{3x}\)

\(A=\frac{x\left(x-1\right)}{3x}\)

\(A=\frac{x-1}{3}\)

b) Thay x = 4 ta có :

\(A=\frac{4-1}{3}=\frac{3}{3}=1\)

c) Để A thuộc Z thì \(x-1⋮3\)

\(\Rightarrow x-1\in B\left(3\right)=\left\{0;3;6;...\right\}\)

\(\Rightarrow x\in\left\{1;4;7;...\right\}\)

Vậy.....

Cho Bt 

a,Tìm điều kiện xác định và rút gọn bt A

b,Tính giá trị bt A tại x=4

c,tìm x thuộc Z để a thuộc Z

điều kiện \(x\ge0\)và x khác 1/4

Q= \(\frac{3\sqrt{x}+2}{2\sqrt{x}-1}+\frac{\sqrt{x}-1}{\sqrt{x}+4}-\frac{x-6\sqrt{x}+5}{2x+7\sqrt{x}-4}=\frac{3x+14\sqrt{x}+8+2x-3\sqrt{x}+1-x+6\sqrt{x}-5}{2x+7\sqrt{x}-4}\)

=\(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}\)

đề Q>1/2 thì \(\frac{4x+17\sqrt{x}+4}{2x+7\sqrt{x}-4}>\frac{1}{2}\)

<=> \(8x+34\sqrt{x}+8>2x+7\sqrt{x}-4\)<=> \(6x+27\sqrt{x}+12>0\) với mọi x>=0

vậy Q>1/2 khi x>=0 và x khác 1/4

đè hinh như là 6\(\sqrt{x}\) nhi bạn