\(A=\dfrac{2x\left(x+1\right)\left(x-2\right)^2}{x\left(x-2\right)\left(x+2\right)\left(x+1\right)}=\dfrac{2\left(x-2\right)}{x+2}\\ A=\dfrac{2\left(\dfrac{1}{2}-2\right)}{\dfrac{1}{2}+2}=\dfrac{2\left(-\dfrac{3}{2}\right)}{\dfrac{5}{2}}=\left(-3\right)\cdot\dfrac{2}{5}=-\dfrac{6}{5}\)

\(B=\dfrac{x\left(x^2-xy+y^2\right)}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\dfrac{x}{x+y}=\dfrac{-5}{-5+10}=\dfrac{-5}{5}=-1\)

a: \(2x\left(2x-1\right)^2-3x\left(x+3\right)\left(x-3\right)-4x\left(x+1\right)^2\)

\(=2x\left(4x^2-4x+1\right)-3x\left(x^2-9\right)-4x\left(x^2+2x+1\right)\)

\(=8x^3-8x^2+2x-3x^3+27x-4x^3-8x^2-4x\)

\(=x^3-16x^2+25x\)

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

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

Tại   x = \(\frac{1}{2}\)thì:

             A = \(\frac{2.\frac{1}{2}-4}{\frac{1}{2}+2}\)\(=\)\(\frac{-3}{\frac{5}{2}}\)\(=\)\(\frac{-6}{5}\)

a: \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

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

\(=\left(2x+1+2x-1\right)^2=\left(4x\right)^2=16x^2\)

b: \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^3+2x^2-x-2-x^3+8\)

\(=2x^2-x+6\)

a) \(\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

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

\(=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2\)

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

\(=\left(4x\right)^2\)

\(=16x^2\)

b) \(\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=\left(x^3+2x^2-x-2\right)-\left(x^3-8\right)\)

\(=x^3+2x^2-x-2-x^3+8\)

\(=2x^2-x+6\)

\(a,\left(2x+1\right)^2+2\left(4x^2-1\right)+\left(2x-1\right)^2\)

\(=\left(2x+1\right)^2+2\cdot\left[\left(2x\right)^2-1^2\right]+\left(2x-1\right)^2\)

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

\(=\left[\left(2x+1\right)+\left(2x-1\right)\right]^2\)

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

\(=\left(4x\right)^2\)

\(=16x^2\)

\(---\)

\(b,\left(x^2-1\right)\left(x+2\right)-\left(x-2\right)\left(x^2+2x+4\right)\)

\(=x^2\left(x+2\right)-\left(x+2\right)-\left(x^3-2^3\right)\)

\(=x^3+2x^2-x-2-x^3+8\)

\(=2x^2-x+6\)

\(\frac{\left(2x^3+2x\right)\left(x-2\right)^2}{\left(x^3-4x\right)\left(x+1\right)}\)

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

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

Thay x=\(\frac{1}{2}\)

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

\(=-1\)

b) \(4x-\sqrt{8}+\frac{\sqrt{x^3+2x^2}}{\sqrt{x+2}}\)

\(=4x-\sqrt{8}+\frac{\sqrt{x^2\left(x+2\right)}}{x+2}\)

\(=4x-\sqrt{8}+\frac{x\left(x+2\right)}{x+2}\)

\(=4x-\sqrt{8}+x\)

\(=5x-\sqrt{8}\)

Với \(x=-\sqrt{2}\) ta có:

  \(5x-\sqrt{8}=5\cdot\left(-\sqrt{2}\right)-\sqrt{4\cdot2}=-5\sqrt{2}-2\sqrt{2}=-7\sqrt{2}\)