a) Ta có: \(\dfrac{2x^2-2x}{x-1}\)

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

=2x

b) Ta có: \(\dfrac{x^2+2x+1}{3x^2+3x}\)

\(=\dfrac{\left(x+1\right)^2}{3x\left(x+1\right)}\)

\(=\dfrac{x+1}{3x}\)

c) Ta có: \(\dfrac{x}{3x-3}+\dfrac{1}{x^2-1}\)

\(=\dfrac{x}{3\left(x-1\right)}+\dfrac{1}{\left(x-1\right)\left(x+1\right)}\)

\(=\dfrac{x+1+3}{3\left(x-1\right)\left(x+1\right)}\)

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

a, \(\dfrac{2x^2-2x}{x-1}=\dfrac{2x\left(x-1\right)}{x-1}=2x\) ( đk : \(x\ne1\) )

b,\(\dfrac{x^2+2x+1}{3x^2+3x}=\dfrac{\left(x+1\right)^2}{3x\left(x+1\right)}=\dfrac{x+1}{3x}\) ( đk : \(x\ne-1\) )

c

=

a: Sửa đề: \(M=\frac{x}{x+3}+\frac{3-x}{x+3}\cdot\frac{x^2+3x+9}{x^2-9}\)

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

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

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

b: \(A=\left(\frac{3x}{1-3x}-\frac{2x}{3x+1}\right):\frac{6x^2+10x}{1-6x+9x^2}\)

\(=\frac{-3x\left(3x+1\right)-2x\left(3x-1\right)}{\left(3x-1\right)\left(3x+1\right)}\cdot\frac{\left(3x-1\right)^2}{2x\left(3x+5\right)}\)

\(=\frac{-9x^2-3x-6x^2+2x}{3x+1}\cdot\frac{3x-1}{2x\left(3x+5\right)}=\frac{-15x^2-x}{3x+1}\cdot\frac{3x-1}{2x\left(3x+5\right)}\)

\(=\frac{-x\left(15x+1\right)}{3x+1}\cdot\frac{3x-1}{2x\left(3x+5\right)}=\frac{-\left(15x+1\right)\left(3x-1\right)}{2\left(3x+1\right)\left(3x+5\right)}\)

\(A=\left(\dfrac{x-1}{x\left(x-2\right)}+\dfrac{x+1}{x\left(x+2\right)}-\dfrac{4}{x\left(x-2\right)\left(x+2\right)}\right)\cdot\dfrac{x\left(x-3\right)}{2\left(x+2\right)}\)

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

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

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

6:

a: ĐKXĐ: x<>0

\(\dfrac{x^3+3x^2+3x+1}{x^2+x}\)

\(=\dfrac{\left(x+1\right)^3}{x\left(x+1\right)}=\dfrac{\left(x+1\right)^2}{x}\)

b: ĐKXĐ: x<>1

\(\dfrac{x^3-3x^2+3x-1}{2x-2}\)

\(=\dfrac{\left(x-1\right)^3}{2\left(x-1\right)}=\dfrac{\left(x-1\right)^2}{2}\)

c: ĐKXĐ: x<>-2

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

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

\(=\dfrac{x+2}{2}\)

d: ĐKXĐ: x<>-2

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

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

e: ĐKXĐ: x<>-y

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

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

\(\dfrac{-3x^2-6x}{4-x^2}=\dfrac{3x^2+6x}{x^2-4}\)

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

7:

a: \(\dfrac{2}{5x^3y^2}=\dfrac{2\cdot4}{20x^3y^2}=\dfrac{8}{20x^3y^2}\)

\(\dfrac{3}{4xy}=\dfrac{3\cdot5\cdot x^2y}{20x^3y^2}=\dfrac{15x^2y}{20x^3y^2}\)

b: \(\dfrac{x}{x^2-2xy+y^2}=\dfrac{x}{\left(x-y\right)^2}\)

\(\dfrac{x}{x^2-xy}=\dfrac{x}{x\left(x-y\right)}=\dfrac{1}{x-y}=\dfrac{\left(x-y\right)}{\left(x-y\right)^2}\)

c: \(\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

\(\dfrac{2}{2x+4}=\dfrac{2}{2\left(x+2\right)}=\dfrac{1}{x+2}=\dfrac{6}{6\left(x+2\right)}\)

\(\dfrac{3}{3x+6}=\dfrac{3}{3\left(x+2\right)}=\dfrac{6}{6\left(x+2\right)}\)

d:

\(\dfrac{2}{2x-6}=\dfrac{2}{2\left(x-3\right)}=\dfrac{1}{x-3};\dfrac{3}{3x-9}=\dfrac{3}{3\left(x-3\right)}=\dfrac{1}{x-3}\)

\(\dfrac{2}{2x-6}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

\(\dfrac{3}{3x-9}=\dfrac{1}{x-3}=\dfrac{x+3}{\left(x-3\right)\left(x+3\right)}\)

\(\dfrac{1}{x+3}=\dfrac{x-3}{\left(x+3\right)\left(x-3\right)}\)

` @ \color{Red}{m}`

` \color{lightblue}{Answer}`  

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

__

\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\\ =\dfrac{3}{2\left(x+3\right)}-\dfrac{x-3}{x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2\left(x-3\right)}{2x\left(x+3\right)}\\ =\dfrac{3x}{2x\left(x+3\right)}-\dfrac{2x-6}{2x\left(x+3\right)}\\ =\dfrac{3x-\left(2x-6\right)}{2x\left(x+3\right)}\\ =\dfrac{3x-2x+6}{2x\left(x+3\right)}\\ =\dfrac{x+6}{2x\left(x+3\right)}\)

__

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

\(\dfrac{x^2}{x^2-1}+\dfrac{x}{\left(1-x\right)\left(x+1\right)}\left(dkxd:x\ne\pm1\right)\)

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

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

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

\(=\dfrac{x}{x+1}\)

========================

\(\dfrac{3}{2x+6}-\dfrac{x-3}{x^2+3x}\left(dkxd:x\ne\pm3;x\ne0\right)\)

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

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

\(=\dfrac{3x-2x+6}{2x\left(x+3\right)}\)

\(=\dfrac{x+6}{2x^2+6x}\)

==========================

\(\dfrac{1}{1-x}+\dfrac{2x}{x^2-1}\left(dkxd:x\ne\pm1\right)\)

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

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

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

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

\(=\dfrac{1}{x+1}\)

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

\(=\dfrac{2x+1-x^2+2x^2-3x-1}{x\left(2x+1\right)}\)

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

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

\(=\dfrac{2x+1}{x\left(2x+1\right)}-\dfrac{x^2}{x\left(2x+1\right)}+\dfrac{2x^2-3x-1}{x\left(2x+1\right)}\)

\(=\dfrac{2x+1-x^2+2x^2-3x-1}{x\left(2x+1\right)}\)

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

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

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

a) Ta có: \(A=\left(\dfrac{\left(x+1\right)^2}{\left(x+1\right)^2-3x}-\dfrac{2x^2+4x-1}{x^3+1}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)

\(=\left(\dfrac{\left(x+1\right)^2}{x^2-x+1}-\dfrac{2x^2+4x-1}{\left(x+1\right)\left(x^2-x+1\right)}-\dfrac{1}{x+1}\right):\dfrac{x^2-4}{3x^2+6x}\)

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

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

\(=\dfrac{x^3+1}{\left(x+1\right)\left(x^2-x+1\right)}\cdot\dfrac{3x}{x-2}\)

\(=\dfrac{3x}{x-2}\)

b) Để A nguyên thì \(3x⋮x-2\)

\(\Leftrightarrow3x-6+6⋮x-2\)

mà \(3x-6⋮x-2\)

nên \(6⋮x-2\)

\(\Leftrightarrow x-2\inƯ\left(6\right)\)

\(\Leftrightarrow x-2\in\left\{1;-1;2;-2;3;-3;6;-6\right\}\)

hay \(x\in\left\{3;1;4;0;5;-1;8;-4\right\}\)

Kết hợp ĐKXĐ, ta được:

\(x\in\left\{3;1;4;0;5;8;-4\right\}\)

Vậy: Để A nguyên thì \(x\in\left\{3;1;4;0;5;8;-4\right\}\)

\(C=\left\lbrack\frac{1}{1+x}+\frac{2x}{1-x^2}\right\rbrack:\left(\frac{1}{x}-1\right)\)

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

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

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

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

\(E=\frac{\left|x-3\right|}{x^2-9}\left(x^2-6x+9\right)\)

\(=\frac{\left|x-3\right|}{\left.\left(x-3\right)\left(x+3\right)\right.}\cdot\left(x-3\right)^2=\frac{\left|x-3\right|\cdot\left(x-3\right)}{x+3}\)

\(F=\frac{\sqrt{x}}{\sqrt{x}-5}-\frac{10\sqrt{x}}{x-25}-\frac{5}{\sqrt{x}+5}\)

\(=\frac{\sqrt{x}\left(\sqrt{x}+5\right)-10\sqrt{x}-5\left(\sqrt{x}-5\right)}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{x-5\sqrt{x}-5\sqrt{x}+25}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}=\frac{\left(\sqrt{x}-5\right)^2}{\left(\sqrt{x}-5\right)\left(\sqrt{x}+5\right)}\)

\(=\frac{\sqrt{x}-5}{\sqrt{x}+5}\)