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

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) \(A=\dfrac{x^2-4x+4}{5x-10}.\) 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) \(Thay\) \(x=-2018:\) \(\dfrac{-2018-2}{5}=-404.\)

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) A đc xác định <=>2x+4\(\left\{{}\begin{matrix}2x+4\ne0\\x^2-4\ne0\end{matrix}\right.\Leftrightarrow\left\{{}\begin{matrix}x\ne-2\\x\ne2\end{matrix}\right.\)

câu b bn quy đòng mẫu là đc

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

b) Ta có: \(A=\dfrac{x}{2x+4}+\dfrac{3x+2}{x^2-4}\)

\(=\dfrac{x}{2\left(x+2\right)}+\dfrac{3x+2}{\left(x-2\right)\left(x+2\right)}\)

\(=\dfrac{x\left(x-2\right)}{2\left(x+2\right)\left(x-2\right)}+\dfrac{2\left(3x+2\right)}{2\left(x-2\right)\left(x+2\right)}\)

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

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

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

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

c) Để A=0 thì \(\dfrac{x+2}{2\left(x-2\right)}=0\)

\(\Leftrightarrow x+2=0\)

hay x=-2(Không thỏa mãn ĐKXĐ)

Vậy: Không có giá trị nào của x để A=0

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

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

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

a) ĐK: \(x\ne4,x\ne2;x\ne-2\)

b) \(A=\dfrac{x^3}{x-4}-\dfrac{x}{x-2}-\dfrac{2}{x+2}\)

\(A=\dfrac{x^3}{\left(x+2\right)\left(x-2\right)}-\dfrac{x\left(x+2\right)}{\left(x+2\right)\left(x-2\right)}-\dfrac{2\left(x-2\right)}{\left(x+2\right)\left(x-2\right)}\)

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

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

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

\(A=\dfrac{\left(x-1\right)\left(x^2-4\right)}{x^2-4}\)

\(A=x-1\)

c) \(A=0\) khi:

\(x-1=0\)

\(\Leftrightarrow x=1\left(tm\right)\)

d) A dương khi: \(A>0\)

\(x-1>0\)

\(\Leftrightarrow x>1\)

Kết hợp với đk: 

\(x>1,x\ne4,x\ne2\)

\(a,ĐK:x\ne0;x\ne1;x\ne\pm2\\ b,A=\left[\dfrac{2+x}{2-x}-\dfrac{2-x}{2+x}+\dfrac{4x^2}{\left(2-x\right)\left(x+2\right)}\right]\cdot\dfrac{x\left(2-x\right)}{x-1}\\ A=\dfrac{x^2+4x+4-x^2+4x-4+4x^2}{\left(2-x\right)\left(x+2\right)}\cdot\dfrac{x\left(2-x\right)}{x-1}\\ A=\dfrac{4x\left(x+1\right)\cdot x}{\left(x+2\right)\left(x-1\right)}=\dfrac{4x^2}{x+2}\)