A xác định khi 5x-10 ≠0 <=> X ≠ 2b) A = x²-4x+4/5x-10= (x-2)²/5(x-2)= x-2/5c) x= -2018<=> A = -2018-2/5= -2020/5 = -404

Chúc bạn học tốt

a) ĐKXĐ: \(x\ne2\)

b) Ta có: \(A=\dfrac{x^2-4x+4}{5x-10}\)

\(=\dfrac{\left(x-2\right)^2}{5\left(x-2\right)}\)

\(=\dfrac{x-2}{5}\)

a: ĐKXĐ: x<>4; x<>-4

b: \(A=\dfrac{\left(x-4\right)\left(x-1\right)}{\left(x-4\right)\left(x+4\right)}=\dfrac{x-1}{x+4}\)

c: Để A nguyên thì x+4-5 chia hết cho x+4

=>\(x+4\in\left\{1;-1;5;-5\right\}\)

=>\(x\in\left\{-3;-5;1;-9\right\}\)

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

a: ĐKXĐ: \(x\notin\left\{2;-2\right\}\)

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

Lời giải:
a.

ĐKXĐ: $x\neq \pm 2$

b.
\(P=\left[\frac{4(x-2)}{(x+2)(x-2)}+\frac{3(x+2)}{(x+2)(x-2)}-\frac{5x+2}{(x-2)(x+2)}\right].\frac{x+2}{2}\)

\(=\frac{4(x-2)+3(x+2)-(5x+2)}{(x-2)(x+2)}.\frac{x+2}{2}=\frac{2(x-2)}{(x-2)(x+2)}.\frac{x+2}{2}=1\)

a: ĐKXĐ: x<>1/2; x<>-1/2; x<>0

b: \(A=\dfrac{4x^2+4x+1-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{5\left(2x-1\right)}{4x}\)

\(=\dfrac{8x}{4x}\cdot\dfrac{5}{2x+1}=\dfrac{10}{2x+1}\)

tự làm đi

dễ vậy cũng hỏi

a: ĐKXĐ: \(x\notin\left\{-\dfrac{1}{2};\dfrac{1}{2};-2\right\}\)

b: \(B=\dfrac{4x^2+4x+1-4-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\cdot\dfrac{2x+1}{x+2}\)

\(=\dfrac{8x-4}{2x-1}\cdot\dfrac{1}{x+2}=\dfrac{4}{x+2}\)