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

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

b: Khi x=3 thì \(A=\dfrac{-3}{3^2+1}=-\dfrac{3}{10}\)

c: x^2+1>=0

=>3/x^2+1>=0

=>-3/x^2+1<=0

=>A<=0(ĐPCM)

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

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

\(b,=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\left(x\ne-y\right)\\ c,=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\left(x\ne-1;x\ne\pm2;x\ne0\right)\)

b: \(\dfrac{x^3-x^2y+xy^2}{x^3+y^3}=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}\)

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

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

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

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

\(=\dfrac{1}{x-y}\)

B) Ta có: 2x-2y-x²+2xy-y²

⇔ 2(x-y)-(x²-2xy+y²)

⇔ 2(x-y)-(x-y)²

⇔ (x-y)(2-x+y)

Đúng thì tick nhé

\(\dfrac{x^3+64}{2x^3-8x^2+32x}\\ =\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}\\ =\dfrac{x+4}{2x}\)

\(\dfrac{x^3+64}{2x^3-8x^3+32x}\)

\(=\dfrac{\left(x+4\right)\left(x^2-4x+16\right)}{2x\left(x^2-4x+16\right)}\)

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