a: \(\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{x+y}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)\)

\(=\frac{x^2+y^2}{x^2y^2}+\frac{2}{x+y}\cdot\frac{x+y}{xy}=\frac{x^2+y^2}{x^2y^2}+\frac{2}{xy}=\frac{x^2+y^2+2xy}{x^2y^2}=\frac{\left(x+y\right)^2}{x^2y^2}\)

\(\left\lbrack\frac{1}{x^2}+\frac{1}{y^2}+\frac{2}{x+y}\cdot\left(\frac{1}{x}+\frac{1}{y}\right)\right\rbrack:\frac{x^3+y^3}{x^2y^2}\)

\(=\frac{\left(x+y\right)^2}{x^2y^2}\cdot\frac{x^2y^2}{\left(x+y\right)\left(x^2-xy+y^2\right)}=\frac{x+y}{x^2-xy+y^2}\)

b: \(\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\)

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

\(=\frac{\left(2x+y+2x-y\right)^2}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}=\frac{16x^2}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}\)

\(\left\lbrack\frac{1}{\left(2x-y\right)^2}+\frac{2}{4x^2-y^2}+\frac{1}{\left(2x+y\right)^2}\right\rbrack\cdot\frac{4x^2+4xy+y^2}{16x}\)

\(=\frac{16x^2}{\left(2x+y\right)^2\cdot\left(2x-y\right)^2}\cdot\frac{\left(2x+y\right)^2}{16x}=\frac{x}{\left(2x-y\right)^2}\)

\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2x^2y+2xy^2\)

\(=x^2\left(x+y\right)+y^2\left(x+y\right)+2xy\left(x+y\right)\)

\(=\left(x+y\right)\left(x^2+y^2+2xy\right)\)

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

\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2xy\left(x+y\right)\)

\(\Leftrightarrow A=\left(x+y\right)\left(x^2+2xy+y^2\right)=\left(x+y\right)\left(x+y\right)^2=\left(x+y\right)^3\)

\(A=x^2\left(x+y\right)+y^2\left(x+y\right)+2x^2y+2xy^2\)

\(\Leftrightarrow A=\left(x^2+y^2\right)\left(x+y\right)+2xy\left(x+y\right)\)

\(\Leftrightarrow\left(x^2+2xy+y^2\right)\left(x+y\right)\)

\(\Leftrightarrow A=\left(x+y\right)^2\left(x+y\right)\)

\(\Leftrightarrow A=\left(x+y\right)^3\)

a) (x + 3)(x² – 3x + 9) – (54 + x³) = (x + 3)(x² – 3x + 3² ) - (54 + x³)

= x³ + 3³ - (54 + x³)

= x³ + 27 - 54 - x³

= -27

b) (2x + y)(4x² – 2xy + y²) – (2x – y)(4x² + 2xy + y²)

= (2x + y)[(2x)² – 2 . x . y + y²] – (2x – y)(2x)² + 2 . x . y + y²]

= [(2x)³ + y³]- [(2x)³ - y³]

= (2x)³ + y³- (2x)³ + y³= 2y³

Bài giải:

a) (x + 3)(x² – 3x + 9) – (54 + x³) = (x + 3)(x² – 3x + 3² ) - (54 + x³)

= x³ + 3³ - (54 + x³)

= x³ + 27 - 54 - x³

= -27

b) (2x + y)(4x² – 2xy + y²) – (2x – y)(4x² + 2xy + y²)

= (2x + y)[(2x)² – 2 . x . y + y²] – (2x – y)(2x)² + 2 . x . y + y²]

= [(2x)³ + y³]- [(2x)³ - y³]

= (2x)³ + y³- (2x)³ + y³= 2y³

a)

\(\begin{array}{l}\left( {2x - 5y} \right)\left( {2x + 5y} \right) + {\left( {2x + 5y} \right)^2}\\ = \left( {2x + 5y} \right)\left( {2x - 5y + 2x + 5y} \right)\\ = \left( {2x + 5y} \right).4x\\ = 2x.4x + 5y.4x\\ = 8{x^2} + 20xy\end{array}\)

b)

\(\begin{array}{l}\left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right) + \left( {2x - y} \right)\left( {4{x^2} + 2xy + {y^2}} \right)\\ = {x^3} + {\left( {2y} \right)^3} + {\left( {2x} \right)^3} - {y^3}\\ = {x^3} + 8{y^3} + 8{x^3} - {y^3}\\ = \left( {{x^3} + 8{x^3}} \right) + \left( {8{y^3} - {y^3}} \right)\\ = 9{x^3} + 7{y^3}\end{array}\)

\(\begin{array}{l}\left( {x - 2y} \right)\left( {{x^2} + 2xy + 4{y^2}} \right) + \left( {x + 2y} \right)\left( {{x^2} - 2xy + 4{y^2}} \right)\\ = {x^3} - {\left( {2y} \right)^3} + {x^3} + {\left( {2y} \right)^3}\\ = {x^3} - 8{y^3} + {x^3} + 8{y^3}\\ = 2{x^3}\end{array}\)

a) \(\left(x-5\right)\left(a^2+5a+25\right)\)

\(=a^3-5^3\)

\(=a^3-125\)

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

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

\(=x^3+8y^3\)