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

a: \(A=\left(\dfrac{x}{\left(x-2\right)\left(x+2\right)}-\dfrac{2}{x-2}+\dfrac{1}{x+2}\right)\cdot\dfrac{x+2}{6}\)

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

b: x=2 ko thỏa mãn ĐKXĐ

=>Loại

Khi x=3 thì A=-1/(3-2)=-1

c: A=2

=>x-2=-1/2

=>x=3/2

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

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

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

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

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

Tick hộ nha

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

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

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

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

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

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

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

b: Để A là số nguyên thì \(x^2+1⋮x^2+x+1\)

=>\(x^2+x+1-x⋮x^2+x+1\)

=>\(x⋮x^2+x+1\)

=>\(x^2+x⋮x^2+x+1\)

=>\(x^2+x+1-1⋮x^2+x+1\)

=>\(-1⋮x^2+x+1\)

=>\(x^2+x+1\in\left\{1;-1\right\}\)

=>\(x^2+x+1=1\)

=>x²+x=0

=>x(x+1)=0

=>\(x\in\left\{0;-1\right\}\)

Bài 2:

a: \(=2x^4-x^3-10x^2-2x^3+x^2+10x=2x^3-3x^3-9x^2+10x\)

b: \(=\left(x^2-15x\right)\left(x^2-7x+3\right)\)

\(=x^4-7x^3+3x^2-15x^3+105x^2-45x\)

\(=x^4-22x^3+108x^2-45x\)

c: \(=12x^5-18x^4+30x^3-24x^2\)

d: \(=-3x^6+2.4x^5-1.2x^4+1.8x^2\)

`a,(25xy^3(2x-y)^2)/(75xy^2(y-2x))(x,y ne 0)(y ne 2x)`

`=(25xy^3(y-2x)^2)/(75xy^2(y-2x))`

`=(y(y-2x))/3`

`b,(x^2-y^2)/(x^2-y^2+xz-yz)`

`=((x-y)(x+y))/((x-y)(x+y)+z(x-y))`

`=(x+y)/(x+y+z)`

`c,((2x+3)-x^2)/(x^2-1)(x ne +-1)`

`=(-(x^2-3x+x-3))/((x-1)(x+1))`

`=(-x(x-3)+x-3)/((x-1)(x+1))`

`=((x-3)(1-x))/((x-1)(x+1))`

`=(3-x)/(1+x)`

`d,(3x^3-7x^2+5x-1)/(2x^3-x^2-4x+3)`

`=(3x^3-3x^2-4x^2+4x+x-1)/(2x^3-2x^2+x^2-x-3x+3)`

`=(3x^2(x-1)-4x(x-1)+x-1)/(2x^2(x-1)+x(x-1)-3(x-1))`

`=(3x^2-4x+1)/(2x^2+x-3)`

`=(3x^2-3x-x+1)/(2x^2-2x+3x-3)`

`=(3x(x-1)-(x-1))/(2x(x-1)+3(x-1))`

`=(3x-1)/(2x+3)`

a) Ta có: \(\dfrac{25xy^3\cdot\left(2x-y\right)^2}{75xy^2\cdot\left(y-2x\right)}\)

\(=\dfrac{25xy^2\cdot y\cdot\left(y-2x\right)^2}{25xy\cdot y\cdot\left(y-2x\right)\cdot3}\)

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