a) = x(x-1) +1 

x(x-1) = 0 khi x = 0; x=1

còn lại x(x - 1) luôn >0

vậy A(x) >0 với mọi x

b) A(x) vô nghiệm vì A(x) luôn .> 0 (cmt)

nhỉn vào dễ thấy

mẫu chung là (4-x²)x

lấy BT chia cho mẫu ở trên (bằng máy)

ra 4x²-8x

đến đây dễ rồi

Rút gọn P

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

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

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

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

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

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

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

y xđ khi \(x^2-3x+1\ne0\)

       \(\Leftrightarrow\left(x-\dfrac{3}{2}\right)^2-\dfrac{5}{4}\ne0\)

\(\Leftrightarrow\left[{}\begin{matrix}x\ne\dfrac{3+\sqrt{5}}{2}\\x\ne\dfrac{3-\sqrt{5}}{2}\end{matrix}\right.\)

\(Q=\left(\dfrac{3-\sqrt{5}}{2}\right)^3+\left(\dfrac{3+\sqrt{5}}{2}\right)^3+3\left(\dfrac{3-\sqrt{5}}{2}+\dfrac{3+\sqrt{5}}{2}\right)\left(\dfrac{3-\sqrt{5}}{2}.\dfrac{3+\sqrt{5}}{2}\right)-13\left(\dfrac{3-\sqrt{5}}{2}.\dfrac{3+\sqrt{5}}{2}\right)\) \(=\left(\dfrac{3-\sqrt{5}}{2}+\dfrac{3+\sqrt{5}}{2}\right)^3-13\)

\(=3^3-13=27-13=14\)