a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)

a: ĐKXĐ: x<>-1

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

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

c: P=2

=>x^2-2x=2x+2

=>x^2-4x-2=0

=>\(x=2\pm\sqrt{6}\)

a) \(A=\dfrac{x^2-4x+4}{5x-10}.\) ĐK: \(x\ne2.\)

b) \(A=\dfrac{x^2-4x+4}{5x-10}=\dfrac{\left(x-2\right)^2}{5\left(x-2\right)}=\dfrac{x-2}{5}.\)

c) \(Thay\) \(x=-2018:\) \(\dfrac{-2018-2}{5}=-404.\)

\(a,ĐK:x\ne\pm2\\ b,A=\dfrac{x^2+4x+4+x^2-4x+4+16}{2\left(x-2\right)\left(x+2\right)}\\ A=\dfrac{2x^2+32}{2\left(x-2\right)\left(x+2\right)}=\dfrac{x^2+16}{x^2-4}\\ c,A=-3\Leftrightarrow-3x^2+12=x^2+16\\ \Leftrightarrow4x^2=-4\Leftrightarrow x\in\varnothing\)

a, điều kiện xác định: x² - 4 ≠ 0    

                           ⇔ x² ≠ 4

                           ⇔x ≠ 2 và x ≠ -2

b,  A= \(\dfrac{x^2}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\)

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

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

       = \(\dfrac{x^2-4}{x^2-4}\)

       = 1

c, x=1    ⇒ A= \(\dfrac{1^2}{1^2-4}-\dfrac{1}{1-2}+\dfrac{2}{1+2}\)

                    = \(\dfrac{4}{3}\)

a) Điều kiện xác định:
A\(\left\{{}\begin{matrix}x-2\ne0\\x+2\ne0\end{matrix}\right.⇔\left\{{}\begin{matrix}x\ne2\\x\ne-2\end{matrix}\right.\)
b) Rút gọn:
A= \(\dfrac{x^2}{x^2-4}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\).

A=  \(\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x}{x-2}+\dfrac{2}{x+2}\).

A= \(\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x\left(x+2\right)}{\left(x-2\right)\left(x+2\right)}+\dfrac{2\left(x-2\right)}{\left(x-2\right)\left(x+2\right)}\)[do MTC là (x-2)(x+2)].
A=  \(\dfrac{x^2}{\left(x-2\right)\left(x+2\right)}-\dfrac{x^2+2x}{\left(x-2\right)\left(x+2\right)}+\dfrac{2x-4}{\left(x-2\right)\left(x+2\right)}\)

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

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

A= \(\dfrac{-4}{\left(x-2\right)\left(x+2\right)}\)

a. ĐKXĐ: x³ - x \(\ne\)0 <=> x(x² - 1) \(\ne\)0 <=> x \(\ne\)0 và x\(\ne\)\(\pm\)1

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

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

\(=\frac{x+1}{x-1}với\)\(x\ne0\)và \(x\ne\pm1\)

\(c.A=2\Leftrightarrow\frac{x+1}{x-1}=2\)

\(\Leftrightarrow\left(x-1\right).2=x+1\)

\(2x-2=x+1\)

\(x=3\)

a) Giá trị của phân thức A xác định

\(\Leftrightarrow x^3-x\ne0\)

\(\Leftrightarrow x\left(x^2-1\right)\ne0\)

\(\Leftrightarrow\hept{\begin{cases}x\ne0\\x\ne1\\x\ne-1\end{cases}}\)

Vậy với \(x\ne0;x\ne\pm1\)thì giá trị của phân thức A đưcọ xác định.

ĐKXĐ: \(x\ne0;x\ne\pm1\)

b) Ta có :

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

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

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

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

c) A = 2

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

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

\(\Leftrightarrow x+1=2x-2\)

\(\Leftrightarrow x-2x=-1-2\)

\(\Leftrightarrow-x=-3\)

\(\Leftrightarrow x=3\)( Thỏa mãn ĐKXĐ )

Vậy ..............

a: ĐKXĐ: \(x\notin\left\{0;-5\right\}\)