\(A=\dfrac{2x^2\left(3x-4y+2\right)}{x\left(3x+y\right)\left(3x-y\right)}=\dfrac{2x\left(3x-4y+2\right)}{\left(3x+y\right)\left(3x-y\right)}\\ A=\dfrac{2\left(3-8+2\right)}{\left(3+2\right)\left(3-2\right)}=\dfrac{2\left(-3\right)}{5}=\dfrac{-6}{5}\)

Ta có: \(\left(x+5\right)\left(x^2-5x+25\right)-\left(x+3\right)^3+\left(x-2\right)\left(x^2+2x+4\right)-\left(x-1\right)^3\)

\(=x^3+125-x^3-9x^2-27x-27+x^3-8-x^3+3x^2-3x+1\)

\(=-6x^2-30x+91\)

\(\left(3x-2y\right)^3+\left(y+2x\right)^3-\left(4x-5y\right)\left(16x^2+20xy+25y^2\right)\)

\(=27x^3-54x^2y+36xy^2-8y^3+y^3+6xy^2+12x^2y+8x^3-\left(64x^3-125y^3\right)\)

\(=35x^3-42x^2y+42xy^2-7y^3-64x^3+125y^3\)

\(=-29x^3-42x^2y+42xy^2+118y^3\)

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

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

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

\(\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{x^2-1}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left(\frac{x+1}{2\left(x-1\right)}+\frac{3}{\left(x-1\right)\left(x+1\right)}-\frac{x+3}{2\left(x+1\right)}\right)\frac{4x^2-4}{5}\)
\(=\left[\frac{\left(x+1\right)^2}{2\left(x-1\right)\left(x+1\right)}+\frac{6}{2\left(x-1\right)\left(x+1\right)}-\frac{\left(x+3\right)\left(x-1\right)}{2\left(x-1\right)\left(x+1\right)}\right]\frac{4x^2-4}{5}\)
\(=\left(\frac{x^2+2x+1+6-x^2+x-3x+3}{2\left(x-1\right)\left(x+1\right)}\right)\frac{4\left(x^2-1\right)}{5}\)
\(=\frac{10}{2\left(x-1\right) \left(x+1\right)}.\frac{4\left(x-1\right)\left(x+1\right)}{5}\)
\(=4\)
Vậy giá trị của biểu thức là 4

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vậy bạn tính giúp bài phía dưới nha bạn 

\(\left[\frac{\left(2x+1\right)^2-\left(2x-1\right)^2}{\left(2x-1\right)\left(2x+1\right)}\right].\frac{5\left(2x-1\right)}{4x}\)

=\(\left[\frac{4x^2+4x+1-4x^2+4x-1}{\left(2x-1\right)\left(2x+1\right)}\right].\frac{5\left(2x-1\right)}{4x}\)

=\(\frac{8x}{\left(2x-1\right)\left(2x+1\right)}.\frac{5\left(2x-1\right)}{4x}\)

=\(\frac{10}{2x+1}\)

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

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

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

Bài này đã có tại đây:

Cho biểu thức:  \(A=\left(\dfrac{2+x}{2-x}-\dfrac{4x^2}{x^2-4}-\dfrac{2-x}{2+x}\right):\dfrac{x^2-3x}{2x^2-x^3}\)Với ... - Hoc24